Answer:
[tex]1^x[/tex] is always 1, so [tex]1^x-1^x=0[/tex]
es0
dame grasias
si soi nuevo
what is the percentage 33% of $145.00
Answer:
47.85
Step-by-step explanation:
33% of 145
of means multiply
Changing to decimal form
.33* 145
47.85
Step-by-step explanation:
cross multiply, set up ur equation like so
33. x
100 = 145
100x=4785
100
x=47.85
In a bag of skittles 12 are yellow, 10 purple, 8 red, 9green and 22 orange. If 5 were selected from the bag. Calculate using counting technique what is the probability that:
A). exactly 2 are red
B). At most 2 are red
Answer:
a) 11.03% probability that exactly two are red.
b) 98.64% probability that at most 2 are red.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this question, the order in which the skittles are chosen is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this question:
12 + 10 + 8 + 9 + 22 = 61 skittles.
A). exactly 2 are red
Desired outcomes:
2 red, from a set of 8.
3 non-red, from a set of 61 - 8 = 53.
So
[tex]D = C_{8,2}*C_{53,3} = \frac{8!}{2!(8-2)!}*\frac{53!}{3!(53-3)!} = 655928[/tex]
Total outcomes:
Five skittles from a set of 61. So
[tex]T = C_{61,5} = \frac{61!}{5!(61-5)!} = 5949147[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{655928}{5949147} = 0.1103[/tex]
11.03% probability that exactly two are red.
B). At most 2 are red
Desired outcomes:
None red(5 from a set of 53)...
One red(from a set of 8), and four non-read(4 from a set of 53).
Two red(655928), as found in a.
So
[tex]D = C_{53,5} + C_{8,1}*C_{53,4} + 655928 = \frac{53!}{5!48!} + \frac{8!}{1!7!}*\frac{53!}{4!49!} + 655928 = 5868213[/tex]
Total outcomes:
Five skittles from a set of 61. So
[tex]T = C_{61,5} = \frac{61!}{5!(61-5)!} = 5949147[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{5868213}{5949147} = 0.9864[/tex]
98.64% probability that at most 2 are red.
The sum of the two digits of a certain two-digit number equals the square root of the number? What is the number?
Answer:
81
Step-by-step explanation:
2-digit squares are 16, 25, 36, 49, 64, 81.
Their sums of digits, are ...
7, 7, 9, 13, 10, 9
The square 81 has a sum of digits equal to its square root.
The federal government recently granted funds for a special program designed to reduce crime in high-crime areas. A study of the results of the program in eight high-crime areas of Miami, Florida, yielded the following results.
Number of Crimes by Area
A B C D E F G H
before 14 7 4 5 17 12 8 9
after 2 7 3 6 8 13 3 5
Has there been a decrease in the number of crimes since the inauguration of the program? Use the .01 significance level. Estimate the p-value
Answer:
Step-by-step explanation:
Hello!
There was a special program funded, designed to reduce crime in 8 areas of Miami.
The number of crimes per area was recorded before and after the program was established in each area. This is an example of a paired data situation. For each are in Miami you have recorded a pair of values:
X₁: Number of crimes recorded in one of the eight areas of Miami before applying the special program.
X₂: Number of crimes recorded in one of the eight areas of Miami after applying the special program.
Area: (Before; After)
A: (14; 2)
B: (7; 7)
C; (4; 3)
D: (5; 6)
E: (17; 8)
F: (12; 13)
G: (8; 3)
H; (9; 5)
To apply a paired sample test you have to define the variable "difference":
Xd= X₁ - X₂
I'll define it as the difference between the crime rate before the program and after the program.
If the original populations have a normal distribution, we can assume that the variable defined from them will also have a normal distribution.
Xd~N(μd; σd²)
If the crime rate decreased after the special program started, you'd expect the population mean of the difference between the crime rates before and after the program started to be less than zero, symbolically μd<0
The hypotheses are:
H₀: μd≥0
H₁: μd<0
α: 0.01
[tex]t= \frac{X[bar]_d-Mu_d}{\frac{S_d}{\sqrt{n} } } ~~t_{n-1}[/tex]
To calculate the sample mean and standard deviation of the variable difference, you have to calculate the difference between each value of each pair:
A= 14 - 2= 12
B= 7 - 7= 0
C= 4 - 3= 1
D= 5 - 6= -1
E= 17 - 8= 9
F= 12 - 13= -1
G= 8 - 3= 5
H= 9 - 5= 4
∑Xdi= 12 + 0 + 1 + (-1) + 9 + (-1) + 5 + 4= 29
∑Xdi²= 12²+0²+1²+1²+9²+1²+5²4²= 269
X[bar]d= 29/8= 3.625= 3.63
[tex]S_d=\sqrt{\frac{1}{n-1}[sumX_d^2-\frac{(sumX_d)^2}{n} ] } = \sqrt{\frac{1}{7}[269-\frac{29^2}{8} ] } = 4.84[/tex]
[tex]t_{H_0}= \frac{3.63-0}{\frac{4.86}{\sqrt{8} } } = 2.11[/tex]
This test is one-tailed to the left and so is the p-value, under a t with n-1= 8-1=7 degrees of freedom, the probability of obtaining a value as extreme as the calculated value is:
P(t₇≤-2.11)= 0.0364
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis. Then at a 1% significance level, you can conclude that the special program didn't reduce the crime rate in the 8 designated areas of Miami.
I hope it helps!
Using the t-distribution, it is found that since the p-value of the test is 0.048 > 0.01, there is not enough evidence to conclude that there has been a decrease in the number of crimes since the inauguration of the program.
At the null hypothesis, it is tested if there has been no reduction, that is, the subtraction of the mean after by the mean before is at least 0, hence:
[tex]H_0: \mu_A - \mu_B \geq 0[/tex]
At the alternative hypothesis, it is tested if there has been a reduction, that is, the subtraction of the mean after by the mean before is negative, hence:
[tex]H_1: \mu_A - \mu_B < 0[/tex]
For both before and after, the mean, standard deviation of the sample(this is why the t-distribution is used) and sample sizes are given by:
[tex]\mu_B = 9.5, s_B = 4.504, n_B = 8[/tex]
[tex]\mu_A = 5.875, s_A = 3.5632, n_A = 8[/tex]
The standard errors are given by:
[tex]s_A = \frac{3.5632}{\sqrt{8}} = 1.2596[/tex]
[tex]s_B = \frac{4.504}{\sqrt{8}} = 1.5924[/tex]
For the distribution of differences, the mean and standard error are given by:
[tex]\overline{x} = \mu_A - \mu_B = 5.875 - 9.5 = -3.625[/tex]
[tex]s = \sqrt{s_A^2 + s_B^2} = \sqrt{1.2596^2 + 1.5924^2} = 2.0304[/tex]
The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.Hence:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{-3.625 - 0}{2.0304}[/tex]
[tex]t = -1.7854[/tex]
The p-value is found using a t-distribution calculator, with t = -1.7854, 8 + 8 - 2 = 14 df and a left-tailed test with a significance level of 0.01, as we are tested if the mean is less than a value.
Using the calculator, the p-value is given by 0.048.Since the p-value of the test is 0.048 > 0.01, there is not enough evidence to conclude that there has been a decrease in the number of crimes since the inauguration of the program.
You can learn more about the use of the t-distribution to test an hypothesis at https://brainly.com/question/13873630
In a given year, 94 cities in the world had populations of 1 million or more. Fifty years later, 530 cities had populations of 1 million or more. What was the percent increase?
Answer:
The percent increase was of 464%.
Step-by-step explanation:
To find the percent increase, first we find how much the current amount is of the original amount. Then, we subtract the current amount from the original amount.
Percentage of current amount:
We solve this using a rule of three.
The original amount(94 cities), was 100% = 1.
The current amount(530 cities) is x. So
94 cities - 1
530 cities - x
94x = 530
x = 530/94
x = 5.64
5.64 = 564% of the original amount
What was the percent increase?
The current amount is 564%
The original amount is 100%
564 - 100 = 464
The percent increase was of 464%.
Simplify: 19w5+ (-3075)
Enter the original expression if it cannot be
simplified.
Enter the correct answer.
ODA
DONE
A bag of trail mix shrugged 1.625 pounds round 1.625 to the nearest tenth.use the number line for help
Answer:
1.625 when convert it to 1.6 pounds
Step-by-step explanation:
The given bag of mill shrugged 1. 625 pounds.
1.625 ≈ 1.6 poundest to the nearest tenth.
A man claims that his lot is triangular, with one side 450 m long and the adjacent side 200 m long. The
angle opposite one side is 28º. Determine the other side length of this lot to the nearest metre.
Given information:
Side a = 450 m (opposite angle A)Side b = 200 m (opposite angle B)Angle A = 28°We can use the Law of Sines to find angle B:
[tex]$\frac{a}{\sin{A}} = \frac{b}{\sin{B}}[/tex]Substitute the given values:
[tex]$\frac{450}{\sin{28^\circ}} = \frac{200}{\sin{B}}[/tex]Now, solve for angle B:
[tex]$\sin{B} = \frac{200 \times \sin{28^\circ}}{450}[/tex][tex]$\sin{B} \approx 0.208654[/tex][tex]$B \approx \arcsin{0.208654} \approx 12.0435^\circ[/tex]Now that we have angle B, we can find angle C using the fact that the sum of the interior angles in a triangle is always 180°:
Angle C = 180° - 28° - 12.0435° Angle C = 139.9565°Now, we can use the Law of Sines again to find the length of the other side (side c) opposite angle C:
[tex]$\frac{a}{\sin{A}} = \frac{c}{\sin{C}}[/tex]Substitute the given values:
[tex]$\frac{450}{\sin{28^\circ}} = \frac{c}{\sin{139.9565^\circ}}[/tex]Now, solve for side c:
[tex]$c = \frac{450 \times \sin{139.9565^\circ}}{\sin{28^\circ}}[/tex][tex]$c \approx 616.685[/tex]To the nearest meter, the other side length of the triangular lot is approximately 617 m. But, since there is not an option for the answer, the closest option is C. 616 m.
________________________________________________________
Full QuestionAnswer:
289 m or 617 m
Step-by-step explanation:
You want the third side length of a triangle with side lengths 450 m and 200 m, with an angle of 28°.
Solution 1The man's claim does not say which side the given angle is opposite. There are two possibilities. (1) It is opposite the unknown side; (2) it is opposite the side of length 450 m. (No triangle is possible having an angle of 28° opposite the shorter given side.)
If the angle is opposite the unknown side, the law of cosines can be used to find the third side length:
c² = a² + b² - 2ab·cos(C)
c² = 450² +200² -2·450·200·cos(28°) ≈ 83569.43
c ≈ √83569.43 ≈ 289 . . . . meters
The other side length could be 289 meters.
Solution 2The third side could also be figured using the law of sines.
a/sin(A) = b/sin(B) = c/sin(C)
450/sin(28°) = 200/sin(B)
B = arcsin(200/450·sin(28°)) ≈ 12.043°
Then angle C is ...
C = 180° -28° -12.043° = 139.957°
and side 'c' is ...
c = 450·sin(139.957°)/sin(28°) ≈ 617 . . . . meters
The other side length could be 617 meters.
__
Additional comment
The problem tells us "one side" is 450 m, and it tells us the angle opposite "one side" is 28°. If both of the descriptors "one side" are referring to the same side, then Solution 2 is the intended one.
The description can be written in a less ambiguous way. As is, we are not sure that the second use of "one side" is referring to any side in particular. Hence the two possibilities.
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A manufacturer of large appliances must decide which of two​ machines, A and​ B, they want to purchase to perform a specific task in the production process. The goal is to buy the machine that has smaller mean time required to perform the task. The plant supervisor selects 15 machine operators at​ random, and each operator performs the task on each of the two machines. The production times are paired for each worker. A paired​ t-test is to be performed to determine if there is evidence that the population mean time using machine A is less than the population mean time using machine B. The summary statistics for the differences in the times required for the task in minutes​ (machine A​ - machine​ B) for the 15 randomly selected workers are given below.
n=15; xÌ… = -10.9 and s=20.3
What must be true about the population of differences in the times required for the task between machine A and machine B for conclusions from the paired t-test to be valid for the population of differences among all workers?
a. Because of the small sample size of differences in times required between machine A and machine B, the distribution of sample means of the differences cannot be normal.
b. Because there were a total of 30 obervations (15 times from machine A and 15 times from machine B), the distribution of sample means of the differences will be approximately normal by the Central Limit Theorem.
c. Because the sample size is "large" enough, the distribution of differences for all workers will be normal.
d. Because of the small sample size of differences in times required between machine A and machine B, the distribution of differences for all workers must be normal.
Answer:
d. Because of the small sample size of differences in times required between machine A and machine B, the distribution of differences for all workers must be normal.
Step-by-step explanation:
A paired t- test conclusion is said to be valid if one of the assumptions that must be satisfied is that: the distribution of the differences must be normal in most cases for which the sample size is small.
From the given information:
the sample size n = 15 ;which is far less than 30
Therefore;we require the distribution of differences in times required between machine A and machine B for all workers to be normal.
From the first option; it is incorrect because even if the sample size is small; the distribution of sample means of the differences will be normal but in the first option ; it is stated that the differences cannot be normal. That makes the first option to be incorrect.
From the second option; is not correct because the sample size (for differences) is 15 and therefore that is a minimal sample which makes the Central Limit Theorem to be invalid and not applicable here.
From the third option; we all know that the sample size is small and not large since it is lesser than 30.
3 1/3 + (-2 1/4) + 1 5/6=
Answer:
[tex]2\dfrac{11}{12}[/tex]
Step-by-step explanation:
The first step is to ensure that all of the fractions have a common denominator.
[tex]3\dfrac{1}{3}+\left( -2\dfrac{1}{4} \right) + 1\dfrac{5}{6}= \\\\3\dfrac{4}{12} - 2\dfrac{3}{12} +1 \dfrac{10}{12}= \\\\1\dfrac{1}{12}+1\dfrac{10}{12}= \\\\\boxed{2\dfrac{11}{12}}[/tex]
Hope this helps!
Create a word problem on Algebraic Expressions and Measurement (grade 10)
Frank can build a fence in twice the time it would take Sandy. Working together, they can build it in 7 hours. How long will it take each of them to do it alone?
Answer
If Frank and Sandy can build the fence in 7 hours, they must be building
1
7
of the fence every hour.
Now, let the amount of time it takes Sandy be
x
hours so that Frank takes
2
x
hours. Sandy can build
1
x
of the fence every hour and Frank can build
1
2
x
of the fence every hour.
We now have the following equation to solve.
1
x
+
1
2
x
=
1
7
2
+
1
2
x
=
1
7
21
=
2
x
x
=
21
2
=
10.50
Thus, Sandy takes
10.50
hours and Frank takes
21
hours.
Please answer this correctly without making mistakes
Answer:
It would be 131.95 if they want the answer in pi it would be 42 pi
Step-by-step explanation:
Assume that the random variable X is normally distributed, with mean muequals45 and standard deviation sigmaequals10. Compute the probability P(57 > than X less than or = 69).
Answer:
0.1069 = 10.69%
Step-by-step explanation:
Problems of normally distributed samples are solved using 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.
In this question:
[tex]\mu = 45, \sigma = 10[/tex]
Between 57 and 69
This is the pvalue of Z when X = 69 subtracted by the pvalue of Z when X = 57. So
X = 69
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{69 - 45}{10}[/tex]
[tex]Z = 2.4[/tex]
[tex]Z = 2.4[/tex] has a pvalue of 0.9918
X = 57
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57 - 45}{10}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849
0.9918 - 0.8849 = 0.1069 = 10.69%
A manufacturer produces both a deluxe and a standard model of an automatic sander designed for home use. Selling prices obtained from a sample of retail outlets follow. Model Price ($) Model Price ($) Retail Outlet Deluxe Standard Retail Outlet Deluxe Standard 1 39 27 5 40 30 2 39 29 6 39 35 3 46 35 7 35 29 4 38 31 The manufacturer's suggested retail prices for the two models show a $10 price differential. Use a .05 level of significance and test that the mean difference between the prices of the two models is $10. a.Calculate the value of the test statistic (to 2 decimals).
b.What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)?
Answer:
Step-by-step explanation:
The data is incorrect. The correct data is:
Deluxe standard
39 27
39 28
45 35
38 30
40 30
39 34
35 29
Solution:
Deluxe standard difference
39 27 12
39 28 11
45 35 10
38 30 8
40 30 10
39 34 5
35 29 6
a) The mean difference between the selling prices of both models is
xd = (12 + 11 + 10 + 8 + 10 + 5 + 6)/7 = 8.86
Standard deviation = √(summation(x - mean)²/n
n = 7
Summation(x - mean)² = (12 - 8.86)^2 + (11 - 8.86)^2 + (10 - 8.86)^2 + (8 - 8.86)^2 + (10 - 8.86)^2 + (5 - 8.86)^2 + (6 - 8.86)^2 = 40.8572
Standard deviation = √(40.8572/7
sd = 2.42
For the null hypothesis
H0: μd = 10
For the alternative hypothesis
H1: μd ≠ 10
This is a two tailed test.
The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 7 - 1 = 6
2) The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
t = (8.86 - 10)/(2.42/√7)
t = - 1.25
We would determine the probability value by using the t test calculator.
p = 0.26
Since alpha, 0.05 < than the p value, 0.26, then we would fail to reject the null hypothesis.
b) Confidence interval is expressed as
Mean difference ± margin of error
Mean difference = 8.86
Margin of error = z × s/√n
z is the test score for the 95% confidence level and it is determined from the t distribution table.
df = 7 - 1 = 6
From the table, test score = 2.447
Margin of error = 2.447 × 2.42/√7 = 2.24
Confidence interval is 8.86 ± 2.24
A 14 gram sample of a substance that's used to preserve fruit and vegetables has a k-value of 0.1092. Find the substance's half-life, in days. Round your answer to the nearest tenth. N=Noe^-kt
Answer:
t = 6.3
Step-by-step explanation:
N=Noe^(-kt)
No = 14 grams
k = .1092
We want to find t when N = 7 or 1/2 of 14
N=Noe^(-kt)
7 = 14 e ^ (-.1092t)
Divide each side by 14
1/2 = e ^ (-.1092t)
take the natural log of each side
ln (1/2) = ln e ^ (-.1092t)
ln (1/2) = -.1092t
Divide each side by -.1092
ln (1/2)/ -.1092 = t
t≈6.3475
Rounding to the nearest tenth
t = 6.3
Answer: 6.3
Step-by-step explanation:
A production line operation is designed to fill cartons with laundry detergent to a mean weight of 32 ounces. A sample of cartons is periodically selected and weighed to determine whether underfilling or overfilling is occurring. If the sample data lead to a conclusion of underfilling or overfilling, the production line will be shut down and adjusted to obtain proper filling. (a) Choose the null and alternative hypotheses that will help in deciding whether to shut down and adjust the production line. H0: - Select your answer - Ha: - Select your answer - (b) Comment on the conclusion and the decision when H0 cannot be rejected. The input in the box below will not be graded, but may be reviewed and considered by your instructor. (c) Comment on the conclusion and the decision when H0 can
Answer:
See explanation below
Step-by-step explanation:
Given a mean of 32. The claim here is that the mean is 32.
Therefore, the null hypothesis and alternative hypothesis, would be:
H0 : u = 32
Ha: u ≠ 32
b) When we fail to reject null hypothesis, H0. This means that the mean weight, u = 32
Conclusion: There is not enough evidence to conclude that there is overfilling or underfilling.
c) When null hypothesis, H0 is rejected. This means the mean weight, u ≠ 32.
Conclusion: There is enough evidence to conclude that overfilling or underfilling exists
The null hypothesis in the sampling is u = 32 and the alternative is that u isn't equal to 32.
How is the null hypothesis depicted?It should be noted that based on the information, when the null hypothesis is rejected, it implies that the weight is 32.
Also, there's no enough evidence to conclude that there's either overfilling or underfilling. When the null hypothesis is rejected, it means that the mean weight is not equal to 32.
Learn more about sampling on:
https://brainly.com/question/17831271
Write 7.3 as a mixed number.
Answer:
7 3/10
Step-by-step explanation:
7.3=7+0.3= 7+ 3/10= 7 3/10
The average amount of rain per year in Greenville is 49 inches. The standard deviation is 8 inches. Find the probability that next year Greenville will receive the following amount of rainfall. Assume the variable is normally distributed.
A) At most 55 inches of rainB) at least 62 inches of rainC) Between 46 and 55 inches of rainD) How many inches of rain would you consider to be an extremely wet year?
Answer:
The probability that next year Greenville will receive the following amount of rainfall A) At most 55 inches of rain is 0.7734, B) at least 62 inches of rain is 0.0516, C) Between 46 and 55 inches of rain is 0.0387
D Having rainfall of above 65 inches would be considered as an extremely wet year
Step-by-step explanation:
A. To find the probability that next year Greenville will receive at most 55 inches of rain we would have to make the following calculation:
P(X<55)
=P((X-mean)/s<(55-49)/8)
=P(Z<0.75)
=0.7734
The probability that next year Greenville will receive at most 55 inches of rain is 0.7734
B. To find the probability that next year Greenville will receive at most 62 inches of rain we would have to make the following calculation:
P(X>62)
=P(Z>(62-49)/8)
=P(Z>1.63)
=0.0516
The probability that next year Greenville will receive at most 62 inches of rain is 0.0516
C. To find the probability that next year Greenville will receive at Between 46 and 55 inches of rain we would have to make the following calculation:
P(46<X<54)
=P((46-49)/8 <Z<(54-49)/8)
=P(-0.38 <Z< 0.63)
=0.3837
The probability that next year Greenville will receive Between 46 and 55 inches of rain is 0.3837
D. Here consider a value that is more two standard deviations above the mean as an unusual vale. That is:
X=2σ+μ
=2(8)+49
=65
Therefore, having rainfall of above 65 inches would be considered as an extremely wet year.
A right triangle has legs 8 and 15. What is its perimeter?
60, 46, 23 or 40?
Answer:
46
Step-by-step explanation:
Answer:
40
Step-by-step explanation:
8 + 15 + 17 = 40
8 + 15 > 17 15 + 17 > 88 + 17 > 15**Hope this helped!!**
**If I was right then please mark brainliest!!**
Please answer this correctly
Answer:
x = 48
Step-by-step explanation:
Since it's given that these two shapes are similar, you can set up a proportion to solve for x, like so:
[tex]\frac{27}{18} =\frac{72}{x}[/tex]
Cross multiply:
[tex]\frac{27x}{1296}[/tex]
Divide 1296 by 27:
x = 48
find the complete factored form of the polynomial -44a^3 + 20a^6
Answer:
[tex]4a^3(5a^3-11)[/tex]
Step-by-step explanation:
→Take out the GCF (Greatest Common Factor). The GCF is [tex]4a^3[/tex] because both [tex]-44a^3[/tex] and [tex]20a^6[/tex] can be divided by it, like so:
[tex]-44a^3+20a^6[/tex]
[tex]4a^3(5a^3-11)[/tex]
simplify (3a-2b)²-2(3a-2b)(a+2b)+(a+2)²
45% of 40 is 18% of what number?
Answer:
100
Step-by-step explanation:
40 / 100= .4
.4 x 45 = 18
18/18= 1
1x 100= 100
Answer:
100
Step-by-step explanation:
:D
If the Math Olympiad Club consists of 11 students, how many different teams of 3 students can be formed for competitions?
Answer:
165 different teams of 3 students can be formed for competitions
Step-by-step explanation:
Combinations of m elements taken from n in n (m≥n) are called all possible groupings that can be made with the m elements so that:
Not all items fitNo matter the order Elements are not repeatedThat is, a combination is an arrangement of elements where the place or position they occupy within the arrangement does not matter. In a combination it is interesting to form groups and their content.
To calculate the number of combinations, the following expression is applied:
[tex]C=\frac{m!}{n!*(m-n)!}[/tex]
It indicates the combinations of m objects taken from among n objects, where the term "n!" is called "factorial of n" and is the multiplication of all the numbers that go from "n" to 1.
In this case:
n: 3m: 11Replacing:
[tex]C=\frac{11!}{3!*(11-3)!}[/tex]
Solving:
[tex]C=\frac{11!}{3!*8!}[/tex]
being:
3!=3*2*1=68!=8*7*6*5*4*3*2*1=40,32011!=39,916,800So:
[tex]C=\frac{39,916,800}{6*40,320}[/tex]
C= 165
165 different teams of 3 students can be formed for competitions
Answer:
There will be 3 teams of 3 students, and one team of 2 students, so there will be 4 teams with one team one student short, but only 3 teams that can hold 3 students
What’s the correct answer for this?
9.4 units.
Because,
Formula for arc length is 2times pie times radius times angle divided by 360.
Answer:
The answer is option 2.
Step-by-step explanation:
You have to use length or arc formula, Arc = θ/360×2×π×r where θ represents degrees and r is radius. Then substitute the following values into the formula :
[tex]arc = \frac{θ}{360} \times 2 \times \pi \times r[/tex]
Let θ = 45,
Let r = 12,
[tex]arc = \frac{45}{360} \times 2 \times \pi \times 12[/tex]
[tex]arc = \frac{1}{8} \times 24\pi[/tex]
[tex]arc = 9.42 \: units \: (3s.f)[/tex]
Which graph represents the following system of inequalities? Y>5x-1 and then y less than or equal to x+3 please it's for Plato
Answer:
y=3
Step-by-step explanation:
If we use LCM and take 3 plus y it equals x so if x equals x it is then y
Answer:
D
Step-by-step explanation:
I took the Plato Course
can anyone please explain me this,would be appreciated
Brian, a scientist, is writing a research paper on projectile motion. during one of his experiments, he throws a ball from a point marked as point a, with a certain velocity in the horizontal direction. the ball travels a horizontal distance of 0.6 meter in the 1st second, 1.2 meters in the 2nd second, 1.8 meters in the 3rd second, and so on. it hits the ground on the 8th second. brian marks the point where the ball landed as point b. calculate the distance between point a and point b.
Answer:
21.6m
Step-by-step explanation:
Brian throws a ball from point 'a'
The ball travels a distance of:
0.6m in the 1^st second1.2m in the 2^nd second1.8m in the 3^rd second2.4m in the 4^th second3.0m in the 5^th second3.6m in the 6^th second4.2m in the 7^th second4.8m in the 8^th secondThe ball travels a total distance of 0.6m + 1.2m + 1.8m + 2.4m + 3.0m + 3.6m + 4.2m + 4.8m = 21.6m from point 'a' to point 'b'.
An engineering school reports that 52% of its students are male (M), 33% of its students are between the ages of 18 and 20 (A), and that 27% are both male and between the ages of 18 and 20. What is the probability of a random student being chosen who is a female and is not between the ages of 18 and 20?
Answer:
42%
Step-by-step explanation:
Given: P(M) = 0.52, P(A) = 0.33, and P(M and A) = 0.27.
Find: P(not M and not A).
P(not M and not A) = 1 − P(M or A)
P(not M and not A) = 1 − (P(M) + P(A) − P(M and A))
P(not M and not A) = 1 − (0.52 + 0.33 − 0.27)
P(not M and not A) = 1 − 0.58
P(not M and not A) = 0.42
Treating these probabilities as Venn probabilities, it is found that there is a 0.42 = 42% probability of a random student being chosen who is a female and is not between the ages of 18 and 20.
-------------------------
The events are:
Event A: Female.Event B: Not between the ages of 18 and 20.-------------------------
52% of the students are male, thus, 48% are female, and [tex]P(A) = 0.48[/tex].33% are between the ages of 18 and 20, thus, 67% are not between these ages, which means that [tex]P(B) = 0.67[/tex]27% are both male and between these ages, which means that 73% are either female or not between these ages, thus [tex]P(A \cup B) = 0.73[/tex].-------------------------
The probability of a random student being chosen who is a female and is not between the ages of 18 and 20 is given by:
[tex]P(A \cap B) = P(A) + P(B) - P(A \cup B)[/tex]
Inserting the probabilities we found:
[tex]P(A \cap B) = 0.48 + 0.67 - 0.73 = 0.42[/tex]
0.42 = 42% probability of a random student being chosen who is a female and is not between the ages of 18 and 20.
A similar problem is given at https://brainly.com/question/21421475
Suppose that twelve bats was used in the experiment. For each trail, the zoo keeper pointed to one of two "feeders". Suppose that the bats went to the correct feeder one that the zoo keeper pointed at) 9 times. Find the 95% confidence interval for the population proportion of times that the bat would follow the point.
a) 0.505,0.995
b) 0.32, 0.81
c) 046, 091
Answer:
a) 0.505,0.995
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 12, \pi = \frac{9}{12} = 0.75[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 - 1.96\sqrt{\frac{0.75*0.25}{12}} = 0.505[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 + 1.96\sqrt{\frac{0.75*0.25}{12}} = 0.995[/tex]
So the correct answer is:
a) 0.505,0.995
A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 9.1 hours.
A ) 0.1046 B) 0.0069 C ) 0.1285 D ) 0.0046
Answer:
B) 0.0069
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846[/tex]
Find the probability that their mean rebuild time exceeds 9.1 hours.
This is 1 subtracted by the pvalue of Z when X = 9.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{9.1 - 8.4}{0.2846}[/tex]
[tex]Z = 2.46[/tex]
[tex]Z = 2.46[/tex] has a pvalue of 0.9931
1 - 0.9931 = 0.0069
So the answer is B.