Answer:
5286 ft
Step-by-step explanation:
The question has been answered but I guess you need an explanation on how the answer was derived.
Given Parameter
1 mi 72 inch
Required
Convert to feet
To complete this conversion, we need to split the given data into 2.
1 mi and 72 inch
= 1 mi + 72 inch
Then we convert them separately.
From unit of standard conversion;
1 mile (1 mi) is equivalent to 5280 ft
Also from unit of standard conversion
1 inch = 1/12 foot
So, 72 inches will be
=> 72 * 1/12 ft
=> 72/12 ft
=> 6 ft
Now that we have the equivalent of 1 mi and 72 in, in foot.
We can then write the following;
1 mi + 72 inch
=> 5280 ft + 6 ft
=> 5286 ft.
Hence, 1 mi 72 inch is equivalent to 5286 ft
Answer:
Use the drop-down menu to complete the comparison.
1 mi 72 in.
=
5,286 ft
Step-by-step explanation:
Water makes about 3/4 of a person’s weight. If a student weighs 30 lbs., how much of his weight is water?
Mr. Collins and Ms. LaPointe are saving money to buy a new motorcycle. The total amount of money Mr. Collins will save is given by the function p(x) = 62 + 5x. The total amount of money Ms. LaPointe will save is given by the function a(x) = x2 + 38. After how many months, x, will they have the same amount of money saved?
We have been given that Mr. Collins and Ms. LaPointe are saving money to buy a new motorcycle. The total amount of money Mr. Collins will save is given by the function [tex]p(x) = 62+5x[/tex]. The total amount of money Ms. LaPointe will save is given by the function [tex]a(x) = x^2+38[/tex]. We are asked to find number of months when they have the same amount of money saved.
To solve our given problem, we will equate both equations as:
[tex]x^2+38=62+5x[/tex]
[tex]x^2-5x+38=62+5x-5x[/tex]
[tex]x^2-5x+38=62[/tex]
[tex]x^2-5x+38-62=62-62[/tex]
[tex]x^2-5x-24=0[/tex]
[tex]x^2-8x+3x-24=0[/tex]
[tex]x(x-8)+3(x-8)=0[/tex]
[tex](x-8)(x+3)=0[/tex]
[tex](x-8)=0,(x+3)=0[/tex]
[tex]x=8,x=-3[/tex]
Since time cannot be negative, therefore, after 8 months they will have the same amount of money saved.
True or False? The segments shown below could form a triangle.
Answer:
True
Step-by-step explanation:
The rule is that every 2 sides' sum is greater than the other side.
So, let's test it.
9+4>11
9+11>4
11+4>9
it checks out!
If the ratio of similarity between two similar polygons is 4:3.2, this ratio converts to
1:8.
8:1.
4:5.
5:4.
Answer:
5 : 4
Step-by-step explanation:
4: 3.2
Multiply by 10 to get rid of the decimal
4*10 : 3.2* 10
40 : 32
Divide each side by 8 to simplify
40/8 : 32/ 8
5 : 4
Suppose a random variable, x, follows a Poisson distribution. Let μ = 2.5 every minute, find the P(X ≥ 125) over an hour. Round answer to 4 decimal places.
Answer:
P(X ≥ 125) = 0.9812
Step-by-step explanation:
To solve this question, we need to understand the Poisson distribution and the normal distribution.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\lambda[/tex] is the mean in the given interval, which is the same as the variance.
Normal distribution:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
The Poisson can be approximated to the normal, with [tex]\mu = \lambda, \sigma = \sqrt{\lambda}[/tex]
Let μ = 2.5 every minute
This is the mean of the Poisson, so [tex]\lambda = 2.5n[/tex], in which n is the number of minutes.
P(X ≥ 125) over an hour
An hour has 60 minutes, so [tex]n = 60, \lambda = 2.5*60 = 150, \sigma = \sqrt{150} = 12.25[/tex]
Using continuity correction, this is [tex]P(X \geq 125 - 0.5) = P(X \geq 124.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 124.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{124.5 - 150}{12.25}[/tex]
[tex]Z = -2.08[/tex]
[tex]Z = -2.08[/tex] has a pvalue of 0.0188
1 - 0.0188 = 0.9812
So
P(X ≥ 125) = 0.9812
Rectangular pieces of land 50m wide and 120m long partitioned to two parts along diagonal what is the length meter of the fence covering one part of it?
Answer:
a
Step-by-step explanation:
bc I just took the quiz☝
The distribution of the scores on a certain exam is N(40,5), which means that the exam scores are Normally distributed with a mean of 40 and standard deviation of 5. a. Sketch the curve and label, on the x-axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be greater than 50. Shade the region under the Normal curve whose area corresponds to this probability.
Answer:
Step-by-step explanation:
Let x represent the random variable representing the scores in the exam. Given that the scores are normally distributed with a mean of 40 and a standard deviation of 5, the diagram representing the curve and the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations is shown in the attached photo
1 standard deviation = 5
2 standard deviations = 2 × 5 = 10
3 standard deviations = 3 × 5 = 15
1 standard deviation from the mean lies between (40 - 5) and (40 + 5)
2 standard deviations from the mean lies between (40 - 10) and (40 + 10)
3 standard deviations from the mean lies between (40 - 15) and (40 + 15)
b) We would apply the probability for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
From the information given,
µ = 40
σ = 5
the probability that a randomly selected score will be greater than 50 is expressed as
P(x > 50) = 1 - P( ≤ x 50)
For x = 50,
z = (50 - 40)/5 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.98
P(x > 50) = 1 - 0.98 = 0.02
The price of oil, in dollars per barrel, declined last week by 3.5%. If it started the week at 102.00 per barrel, at what per barrel price did it end the week?
Answer:
98.57
Step-by-step explanation:
3.5% over 100% X 102=3.43
102-3.43= 98.57
prove
cos A /(1- sin A) = (1 + sin A)/cos A
Answer:
answer is in exaplation
Step-by-step explanation:
cosA
+
cosA
1+sinA
=2secA
Step-by-step explanation:
\begin{lgathered}LHS = \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=\frac{cos^{2}A+(1+sinA)^{2}}{(1+sinA)cosA}\\=\frac{cos^{2}A+1^{2}+sin^{2}A+2sinA}{(1+sinA)cosA}\\=\frac{(cos^{2}A+sin^{2}A)+1+2sinA}{(1+sinA)cosA}\\=\frac{1+1+2sinA}{(1+sinA)cosA}\end{lgathered}
LHS=
1+sinA
cosA
+
cosA
1+sinA
=
(1+sinA)cosA
cos
2
A+(1+sinA)
2
=
(1+sinA)cosA
cos
2
A+1
2
+sin
2
A+2sinA
=
(1+sinA)cosA
(cos
2
A+sin
2
A)+1+2sinA
=
(1+sinA)cosA
1+1+2sinA
/* By Trigonometric identity:
cos² A+ sin² A = 1 */
\begin{lgathered}=\frac{2+2sinA}{(1+sinA)cosA}\\=\frac{2(1+sinA)}{(1+sinA)cosA}\\\end{lgathered}
=
(1+sinA)cosA
2+2sinA
=
(1+sinA)cosA
2(1+sinA)
After cancellation,we get
\begin{lgathered}= \frac{2}{cosA}\\=2secA\\=RHS\end{lgathered}
=
cosA
2
=2secA
=RHS
Therefore,
\begin{lgathered}\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA\end{lgathered}
1+sinA
cosA
+
cosA
1+sinA
=2secA
A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical -value approach.
= 20.5, n = 11 , σ = 7, H0: μ = 18.7; Ha: μ ≠ 18.7, α = 0.01
Answer:
Z = 0.8528 < 2.576
The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance
Null hypothesis is Accepted at 0.01 level of significance.
There is no significance difference between the means
Step-by-step explanation:
Given data
size of the sample 'n' = 11
mean of the sample x⁻ =20.5
Mean of the Population μ = 18.7
Standard deviation of Population σ = 7
Test statistic
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{20.5 -18.7}{\frac{7}{\sqrt{11} } }[/tex]
[tex]Z = \frac{1.8}{2.1105}[/tex]
Z = 0.8528
critical Value
[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} = 2.576[/tex]
The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance
Null hypothesis is Accepted at 0.01 level of significance.
There is no significance difference between the means
Answer:Answer:
B. 18.7 ± 9.7
Step-by-step explanation:
Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 32 mpg with a standard deviation of 6 mpg. Thirty-one non-hybrid sedans get a mean of 21 mpg with a standard deviation of three mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test at the 5% level to evaluate the manufacturers claim. NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Answer:
There is enough evidence to support the claim of the manufacturers that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.
Step-by-step explanation:
This is a hypothesis test for the difference between populations means.
The claim is that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0[/tex]
The significance level is 0.05.
The sample 1 (hybrid), of size n1=21 has a mean of 32 and a standard deviation of 6.
The sample 2 (non-hybrid), of size n2=31 has a mean of 21 and a standard deviation of 3.
The difference between sample means is Md=11.
[tex]M_d=M_1-M_2=32-21=11[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{6^2}{21}+\dfrac{3^2}{31}}\\\\\\s_{M_d}=\sqrt{1.714+0.29}=\sqrt{2.005}=1.4158[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{11-0}{1.4158}=\dfrac{11}{1.4158}=7.77[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-2=21+31-2=50[/tex]
This test is a right-tailed test, with 50 degrees of freedom and t=7.77, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>7.77)=0.0000000002[/tex]
As the P-value is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.
PPPLLLLLEEAAASSSEEE HELP!!!!!!!!!
Consider the following situation, which involves two options. Determine which option is less expensive. Are there unstated factors that might affect your decision?
You currently drive 275 miles per week in a car that gets 16 miles per gallon of gas. You are considering buying a new fuel-efficient car for $18 comma 000 (after trade-in on your current car) that gets 44 miles per gallon. Insurance premiums for the new and old car are $1000 and $600 per year, respectively. You anticipate spending $1100 per year on repairs for the old car and having no repairs on the new car. Assume gas costs $3.50 per gallon. Over a five-year period, is it less expensive to keep your old car or buy the new car?
Over a five-year period, the cost of the old car is $____ and the cost of the new car is $_____
Thus, over a five-year period, it is less expensive to keep the old car or buy the new car?
Answer:
Step-by-step explanation:
A customer will be charged extra if the weight of their suitcase is above 48 pounds.
Write an inequality to represent w, the weight of the suitcase in pounds, that will have an extra charge.
Enter your answer by clicking and using the expression evaluator.
Answer:
[tex]w>48[/tex] pounds
Step-by-step explanation:
The customer will be charged extra if the weight of their suitcase is above 48 pounds.
Let the weight of the suitcase = w (in pounds)
Therefore, w above (greater than) 48 pounds is written mathematically as:
[tex]w>48[/tex] pounds
This is the inequality that represents w, the weight of the suitcase in pounds, that will have an extra charge.
Please answer this correctly I want genius or expert people to answer this correctly.
Answer:
18.84mm^2
Step-by-step explanation:
Surface area of a cylinder can be found with the following formula:
2πrh+2πr^2
r is radius
h is height
Plug our values in
2π1(2)+2π(1)^2
2π(2)+2π(1)
4π+2π
6π
6(3.14)
18.84
Waiting times (in minutes) of customers at a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. Find the coefficient of variation for each of the two sets of data, then compare the variation. Bank A (single line): 6.5 nbsp 6.7 nbsp 6.7 nbsp 6.8 nbsp 7.1 nbsp 7.3 nbsp 7.4 nbsp 7.7 nbsp 7.7 nbsp 7.7 Bank B (individual lines): 4.3 nbsp 5.4
Complete Question
1. Waiting times (in minutes) of customers at a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. Find the coefficient of variation for each of the two sets of data, then compare the variation.
Bank A (single lines): 6.5, 6.6, 6.7, 6.8, 7.2, 7.3, 7.4, 7.6, 7.6, 7.7
Bank B (individual lines): 4.1, 5.4, 5.8, 6.3, 6.8, 7.8, 7.8, 8.6, 9.3, 9.7
- The coefficient of variation for the waiting times at Bank A is ----- %?
- The coefficient of variation for the waiting times at the Bank B is ----- %?
- Is there a difference in variation between the two data sets?
Answer:
a
The coefficient of variation for the waiting times at Bank A is [tex]l =[/tex]6.3%
b
The coefficient of variation for the waiting times at Bank B is [tex]l_1 =[/tex]25.116%
c
The waiting time of Bank B has a considerable higher variation than that of Bank A
Step-by-step explanation:
From the question we are told that
For Bank A : 6.5, 6.6, 6.7, 6.8, 7.2, 7.3, 7.4, 7.6, 7.6, 7.7
For Bank B : 4.1, 5.4, 5.8, 6.3, 6.8, 7.8, 7.8, 8.6, 9.3, 9.7
The sample size is n =10
The mean for Bank A is
[tex]\mu_A = \frac{6.5+ 6.6+ 6.7+ 6.8+ 7.2+ 7.3+ 7.4+ 7.6+ 7.6+ 7.7}{10}[/tex]
[tex]\mu_A = 7.14[/tex]
The standard deviation is mathematically represented as
[tex]\sigma = \sqrt{\frac{\sum|x- \mu|}{n} }[/tex]
[tex]k = \sum |x- \mu | ^2 = 6.5 -7.14|^2 + |6.6-7.14|^2+ |6.7-7.14|^2+ |6.8-7.14|^2 + |7.2-7.14|^2+ |7.3-7.14|^2, |7.4-7.14|^2+ |7.6-7.14|^2+|7.6-7.14|^2+|7.7-7.14|^2[/tex]
[tex]k = 2.42655[/tex]
[tex]\sigma = \sqrt{\frac{2.42655}{10} }[/tex]
[tex]\sigma = 0.493[/tex]
The coefficient of variation for the waiting times at Bank A is mathematically represented as
[tex]l = \frac{\sigma}{\mu} *100[/tex]
[tex]l = \frac{0.493}{7.14} *100[/tex]
[tex]l =[/tex]6.3%
Considering Bank B
The mean for Bank B is
[tex]\mu_1 = \frac{4.1+ 5.4+ 5.8+ 6.3+6.8+ 7.8+ 7.8+ 8.6+ 9.3+ 9.7}{10}[/tex]
[tex]\mu_1 = 7.16[/tex]
The standard deviation is mathematically represented as
[tex]\sigma_1 = \sqrt{\frac{\sum|x- \mu_1|}{n} }[/tex]
[tex]\sum |x- \mu_1 | ^2 =4.1-7.16|^2 +| 5.4-7.16|^2+ |5.8-7.16|^2 + | 6.3-7.16|^2 + |6.8-7.16|^2 + | 7.8-7.16|^2 +|7.8-7.16|^2 +|8.6-7.16|^2 + |9.3-7.16|^2 +|9.7-7.16|^2[/tex]
[tex]\sum |x- \mu_1 | ^2 =32.34[/tex]
[tex]\sigma_1 = \sqrt{\frac{32.34}{10} }[/tex]
[tex]\sigma_1 = 1.7983[/tex]
The coefficient of variation for the waiting times at Bank B is mathematically represented as
[tex]l_1 = \frac{\sigma }{\mu} *100[/tex]
[tex]l_1 = \frac{1.7983 }{7.16} *100[/tex]
[tex]l_1 =[/tex]25.116%
Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of cigarettes. Construct a 99% confidence interval for the mean nicotine content of this brand of cigarette.
a 27.3 ± 3.033
b 27.3 ± 3.321
c 27.3 ± 3.217
d 27.3 ± 3.131
Part II - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%. Do you agree?
a No, since the value 28.4 does not fall in the 98% confidence interval.
b
Yes, since the value 28.4 does fall in the 98% confidence interval.
c
Yes, since the value 28.4 does not fall in the 98% confidence interval.
d
No, since the value 28.4 does fall in the 98% confidence interval.
Answer:
(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].
(II) No, since the value 28.4 does not fall in the 98% confidence interval.
Step-by-step explanation:
We are given that a new cigarette has recently been marketed.
The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.
Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean nicotine content = 27.3 milligrams
s = sample standard deviation = 2.8 milligrams
n = sample of cigarettes = 9
[tex]\mu[/tex] = true mean nicotine content
Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
Part I : So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-3.355 < [tex]t_8[/tex] < 3.355) = 0.99 {As the critical value of t at 8 degree
of freedom are -3.355 & 3.355 with P = 0.5%}
P(-3.355 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 3.355) = 0.99
P( [tex]-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }[/tex] , [tex]27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }[/tex] ]
= [27.3 [tex]\pm[/tex] 3.131]
= [24.169 mg , 30.431 mg]
Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].
Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.
The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.
The Pivotal quantity for 98% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean nicotine content = 24.9 milligrams
s = sample standard deviation = 2.6 milligrams
n = sample of cigarettes = 9
[tex]\mu[/tex] = true mean nicotine content
Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 98% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.896 < [tex]t_8[/tex] < 2.896) = 0.98 {As the critical value of t at 8 degree
of freedom are -2.896 & 2.896 with P = 1%}
P(-2.896 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.896) = 0.98
P( [tex]-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
P( [tex]\bar X-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
98% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }[/tex] , [tex]24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }[/tex] ]
= [22.4 mg , 27.4 mg]
Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].
No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.
Determine whether the results appear to have statistical significance, and also determine whether the results appear to have practical significance. In a study of a gender selection method used to increase the likelihood of a baby being born a girl, 1961 users of the method gave birth to 961 boys and 1000 girls. There is about a 20% chance of getting that many girls if the method had no effect. 1. Does the weight loss program have statistical significance?A. Yes, the program is statistically significant because the results are unlikely to occur by chance.B. Yes, the program is statistically significant because the results are likely to occur by chance.C. No, the program is not statistically significant because the results are likely to occur by chance.D. No, the program is not statistically significant because the results are unlikely to occur by chance.2. Does the weight loss program have practical significance?A. Yes, the program is practically significant because the results are too unlikely to occur by chance.B. No, the program is not practically significant because the results are likely to occur even if the weight loss program has no effect.C. No, the program is not practically significant because the amount of weight lost is trivial.D. Yes, the program is practically significant because the amount of lost weight is large enough to be considered practically significant.
Answer:
C. No, the program is not statistically significant because the results are likely to occur by chance.
Step-by-step explanation:
At a significance level that is smaller than 0.2, the effect is not significant.
We have a P-value of 20%, which means that we have 20% chances of getting this sample given that the method has no effect.
We then can conclude that there is not enough evidence to support the claim that the method is effective.
The temperature outside when Colin went to bed was -4°F. When he woke up the next
morning, it was -11°F outside. Describe the change in temperature by completing the
statements.
Step-by-step explanation:
The temperature outside when Colin went to bed was -4°F. When he woke up the next morning, it was -11°F outside
To find the change in the temperature , find the difference in temperature
Change in temperature = temperature in morning - temperature at night
change in temperature = [tex]-11 -(-4)= -7[/tex]
temperature changed by -7 °F
So the temperature is dropped by 7°F
Answer:
the temperature is dropped by 7°F
The temperature change when he woke up the next morning given the data was –7 °F.
Data obtained from the questionFrom the question given above , the following data were obtained:
Temperature at night (T₁) = –4 °F Temperature in the morning (T₂) = –11 °F Change in temperature (ΔT) =? How to determine the change in the temperatureThe temperature change when Colin woke up can be obtained by taking the difference in the temperature at night and morning. This is illustrated as follow:
ΔT = T₂ – T₁
ΔT = –11 – (–4)
ΔT = –11 + 4
ΔT = –7 °F
Thus, from the calculation made above, we can conclude that there was a temperature drop of –7 °F when he woke up the next morning
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Please answer this question !! Thank you !! Will give brainliest !! Really important !!
Answer:
0 = 3x -2y -6
Step-by-step explanation:
The general form of the equation of a line is 0 = Ax + By + C
y = 3/2x -3
Subtract y from each side
0 = 3/2x -y -3
Multiply each side by 2
0 = 3x -2y -6
Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if a relation is not an equivalence relation, please list each property that is missing and the reason why it is missing.) (1) {(0,0), (1,1), (2,2), (3,3)} (2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)} (3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)} (4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}
Answer:
(1)Equivalence Relation
(2)Not Transitive, (0,3) is missing
(3)Equivalence Relation
(4)Not symmetric and Not Transitive, (2,1) is not in the set
Step-by-step explanation:
A set is said to be an equivalence relation if it satisfies the following conditions:
Reflexivity: If [tex]\forall x \in A, x \rightarrow x[/tex]Symmetry: [tex]\forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x[/tex]Transitivity: [tex]\forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z[/tex](1) {(0,0), (1,1), (2,2), (3,3)}
(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}
The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.
(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}
In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.
Therefore, It is not transitive.
As a result, the set (2) is not an equivalence relation.
(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}
(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric
Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.
Lines a and b are parallel.
Parallel lines a and b are cut by transversals s and t. The angles formed by the intersection of lines a, s, and t, clockwise from top left are (7 x) degrees, blank; formed by lines b and s are blank, blank, blank, (11 x) degrees; formed by lines b and t are 90 degrees,
What is the value of x?
5
10
35
55
If you answer correctly, we will name the brainiest!
This is a challenge question, you have 5 minutes.
Answer:
The answer is x=5.
Step-by-step explanation:
11x + 7x + 90 = 180
18x + 90 = 180
-90 -90
= 18x = 90
x = 5
x is 5!
Value of x is 10°.
Correct option is B.
What is angle?A figure which is formed by two rays or lines that shares a common endpoint is called an angle. The word “angle” is derived from the Latin word “angulus”, which means “corner”.
Given,
a || b
cut by transversal s and t
lines b and t have angle = 90°
The first top left angle between a, s, t = 7x
The last bottom angle between s and b = 11x
a || b, so a and t will also have angle = 90°
by the figure,
180° - 11x + 90° - 7x + 90° = 180°
360° - 18x = 180°
18x = 180°
x = 10°
Hence, 10° is value of x.
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What is the area of a circle with a radius of 6 inches?
9. in.
O 125 in.
O 36in 2
81 in 2
Answer:
A = 36pi in^2
Step-by-step explanation:
The area of a circle is given by
A = pi r^2
A = pi 6^2
A = 36pi in^2
Radius(r)=6 inches
Area of circle=?
Now,
Area of circle=pi r ^2
=pi (6)^2
36 pi inches^2
So the right answer is 36 pi inches^2
Hope it helps
if the perimeter of a square gardern is 84feet . what is the area of the qardern
Answer:
441
Step-by-step explanation:
84 / 4 = 21
21 x 21 = 441
Answer:
441
Step-by-step explanation:
84/2=42
Which function represents exponential decay? f(x) = f(x) = f(x) = 4 f(x) = 4
Answer:
f(x)=4
Step-by-step explanation:
In mathematics, exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. It can be expressed by the formula y=a(1-b)x wherein y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed.
f(x)=4(2/3)*
option D
opción D
2. 4 masks and 1 pack of gloves costs $18. Three packs of gloves and 4 masks costs $22. What is the cost of one pack of gloves and one mask?
Answer:
One mask=$4
One pack of glove=$2
Step-by-step explanation:
Let mask=m
Let glove=g
4m+g=18 (1)
4m+3g=22 (2)
From (1)
g=18-4m
Substitute g=18-4m into equation (2)
4m+3g=22
4m+3(18-4m)=22
4m+54-12m=22
4m-12m=22-54
-8m=-32
Divide both sides by -8
m=4
Substitute m=4 into equation (1)
4m+g=18
4(4)+g=18
16+g=18
g=18-16
g=2
One mask=$4
One pack of glove=$2
Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the event that the A2 right leg tag is lost. Suppose these two events are independent and P(A1)=P(A2)=0.4. Find the probability that exactly one tag is lost, given that at least one tag is lost (write it up to second decimal place).
Answer:
0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost
Step-by-step explanation:
Independent events:
If two events, A and B, are independent, then:
[tex]P(A \cap B) = P(A)*P(B)[/tex]
Conditional probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: At least one tag is lost
Event B: Exactly one tag is lost.
Each tag has a 40% = 0.4 probability of being lost.
Probability of at least one tag is lost:
Either no tags are lost, or at least one is. The sum of the probabilities of these events is 1. Then
[tex]p + P(A) = 1[/tex]
p is the probability none are lost. Each one has a 60% = 0.6 probability of not being lost, and they are independent. So
p = 0.6*0.6 = 0.36
Then
[tex]P(A) = 1 - p = 1 - 0.36 = 0.64[/tex]
Intersection:
The intersection between at least one lost(A) and exactly one lost(B) is exactly one lost.
Then
Probability at least one lost:
First lost(0.4 probability) and second not lost(0.6 probability)
Or
First not lost(0.6 probability) and second lost(0.4 probability)
So
[tex]P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48[/tex]
Find the probability that exactly one tag is lost, given that at least one tag is lost (write it up to second decimal place).
[tex]P(B|A) = \frac{0.48}{0.64} = 0.75[/tex]
0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost
How to solve? What are the rules in BODMAS
Answer:
18/35
Step-by-step explanation:
1 1/5 ÷ 2 1/3
Change from mixed numbers to improper fractions
1 1/5 = (5*1+1)/5 = 6/5
2 1/3 = (3*2+1)/3 = 7/3
6/5÷ 7/3
Copy dot flip
6/5* 3/7
18/35
Answer:
Step-by-step explanation:
I don't say you must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me my friend...
A population of values has a normal distribution with μ = 247 and σ = 62.2. You intend to draw a random sample of size n = 16. (a) Find the probability that a single randomly selected value is greater than 295.2. (b) Find the probability that a sample of size n= 16 is randomly selected with a mean greater than 295.2. Give your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Answer:
a) [tex] z = \frac{295.2-247}{62.2}=0.772[/tex]
And using the normal distribution table or excel we got:
[tex] P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006[/tex]
b) [tex] z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119[/tex]
And we can use the normal standard table or excel in order to find the probability and we got:
[tex] P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907[/tex]
Step-by-step explanation:
For this case we know that the random variable of interest is normally distributed with the following parameters:
[tex] X \sim N (\mu = 247, \sigma =62.2)[/tex]
Part a
We want to find this probability:
[tex] P(X>295.2)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
Replacing we got:
[tex] z = \frac{295.2-247}{62.2}=0.772[/tex]
And using the normal distribution table or excel we got:
[tex] P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006[/tex]
Part b
We select a random sample of size n = 16 and we try to find this probability:
[tex] P(\bar X >295.2)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119[/tex]
And we can use the normal standard table or excel in order to find the probability and we got:
[tex] P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907[/tex]
A company had a gross profit of $300,000 based on sales of $400,000. Its cost of goods sold equals $700,000 is this true or false?
Answer:
false
Step-by-step explanation:
Answer:
Its true
Step-by-step explanation:
I took the quiz for Intro to Careers in Finance
Choose a reasonable estimate for the amount of water a cup would hold.
Answer:
A cup of water can hold a cup of water (or about 250 mL)
Step-by-step explanation:
anything between 1 and 4 cups should be an acceptable answer
If this helps, please consider giving me brainliest
Answer:
I'm gonna go with 250 ML
Step-by-step explanation:
One liter is a little more than 1 quart.
250 mL = 0.25 L