What’s the correct answer for this?

Answer:

b

Step-by-step explanation:

Answer:

Step-by-step explanation:

The person receives 10000 every month

for one year total principal 10000 * 12 = 120000

r= 12%p.a and t is 1yr

simple interest= 120000*12*1/100 = 14400

total money received in present year 120000 + 14400 = 134400

for 20 yrs principal becomes 10000*12*20= $2400000

simple interest = 2400000*12*20/100= $5760000

total amount received for 20 yrs = 2400000+ 5760000=8160000.

present value of principal as money receive 10000 for 10 yrs = 10000*10 = 100000

si = 100000*10*12/100= 120000

amount = $220000

Answer:-111

Step-by-step explanation:

Answer:

5.029ft

Step-by-step explanation:

[tex](titer \div 360) \times \pi \times diameter[/tex]

Answer:

[tex]4x^2+3x-2[/tex]

Step-by-step explanation:

We are given that

[tex]\frac{8x^3+2x^2-7x+2}{2x-1}[/tex]

We have to divide the polynomials.

We have

Divisor polynomial=[tex]2x-1[/tex]

Dividend polynomial=[tex]8x^3+2x^2-7x+2[/tex]

[tex]\frac{8x^3+2x^2-7x+2}{2x-1}=4x^2+3x-2[/tex]

Quotient=[tex]4x^2+3x-2[/tex]

Remainder=0

Answer:

  300 mL

Step-by-step explanation:

If one part is 100 mL, then 3 parts will be 300 mL.

300 mL of water should be added.

9514 1404 393

Answer:

  55.39 square units

Step-by-step explanation:

The formula for the area of a circle is ...

  A = πr²

Put the given numbers in place of the corresponding variables and do the arithmetic.

  A = 3.14 × 4.2² = 3.14 × 17.64 = 55.3896

Rounded to hundredths, the area is ...

  55.39 square units

Answer:

0.2308 = 23.08% probability that at least 10 but fewer than 13 trucks pass the inspection

Step-by-step explanation:

For each truck, there are only two possible outcomes. Either they pass the inspection, or they do not. The probability of a truck passing the inspection is independent of other trucks. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [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]

And p is the probability of X happening.

80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection.

This means that [tex]p = 0.8[/tex]

17 trucks:

This means that [tex]n = 17[/tex]

What is the probability that at least 10 but fewer than 13 trucks pass the inspection

[tex]P(10 \leq X < 13) = P(X = 10) + P(X = 11) + P(X = 12)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 10) = C_{17,10}.(0.8)^{10}.(0.2)^{7} = 0.0267[/tex]

[tex]P(X = 11) = C_{17,11}.(0.8)^{11}.(0.2)^{6} = 0.0680[/tex]

[tex]P(X = 12) = C_{17,12}.(0.8)^{12}.(0.2)^{5} = 0.1361[/tex]

[tex]P(10 \leq X < 13) = P(X = 10) + P(X = 11) + P(X = 12) = 0.0267 + 0.0680 + 0.1361 = 0.2308[/tex]

0.2308 = 23.08% probability that at least 10 but fewer than 13 trucks pass the inspection

The percent of daily calories does one serving of walnuts provide is 9%.

This question is solved using proportions.

We have:

-----------------

Walnut:

Thus, the amount of calories of fat is:

[tex]A = 20 \times 9 = 180[/tex]

-----------------

Percentage:

180 calories out of a daily diet of 2000 calories, so:

180*100%/2000 = 9%

The percent of daily calories does one serving of walnuts provide is 9%.

A similar question is found at https://brainly.com/question/10491646

Answer:

(c) Observational Distribution

(d) Diet Choice

Step-by-step explanation:

Out of 1033 American adults interviewed in July 2018, 5% consider themselves to be vegetarian.

Since the poll observes the diet habit of the respondents, the cited 5% is part of the Observational distribution of whether the respondent is a vegetarian or not (which is the Diet Choice).

If on the other hand, the poll seeks to manipulate the conditions of the study, it would have been an experimental distribution.

Answer:

no

Step-by-step explanation:

Answer:0.06

Step-by-step explanation:

Answer:

a) [tex]f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000[/tex]

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

[tex]f'(t)=0.001155t(t-1980)^{0.5}[/tex]

a) To find the function f(t) you integrate f(t):

[tex]\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt[/tex]

To solve the integral you use:

[tex]\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}[/tex]

Next, you replace in the integral:

[tex]\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C[/tex]

Then, the function f(t) is:

[tex]f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'[/tex]

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

[tex]f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000[/tex]

b) For t = 2015 you have:

[tex]f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7[/tex]

Answer: D

Step-by-step explanation: y int is when x=0, thus when there are no chirps.

Y=(1/6)x + 50

put in 0

Y=(1/6)*0 + 50

Y=50

Slope is the rate at which x changes, in this case, 1/6

Step-by-step explanation:

here is the answer u can get help from it.....

The solution to the equation (2x-4)/5 = 4 is x = 12.

To solve the equation (2x-4)/5 = 4, we will use the following steps.

Multiply both sides of the equation by 5 to eliminate the denominator as;

5[(2x-4)/5] = 5 x 4

This simplifies to:

2x - 4 = 20

Add 4 to both sides of the equation to isolate the term with x:

2x - 4 + 4 = 20 + 4

This simplifies to:

2x = 24

Divide both sides of the equation by 2 to solve for x:

(2x)/2 = 24/2

This simplifies to:

x = 12

Thus, the solution to the equation (2x-4)/5 = 4 is x = 12.

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#SPJ2

Answer: 82

Step-by-step explanation:

72 times 6 is equal to 432

6 plus 2 is 8

74.25 times 8 is equal to 594

If you minus 594 and 432 you get 162

162 is total marks for both subjects

162 minus 80 you get 82 which is the marks for the 8th subject

Hope it taught you something..

Substitute [tex]y=x-8[/tex] into the second equation to get [tex]2x+3(x-8)=1,[/tex] which simplifies to [tex]5x=25,[/tex] so [tex]\boxed{x=5}.[/tex] It easily follows that [tex]\boxed{y=-3}.[/tex]

Answer:

x=5, y=-3

Step-by-step explanation:

Answer:

Mean: 18.6 years old

Median: 18.5 years old

Step-by-step explanation:

Mean = average (add all numbers then divide by the number of numbers in the data set)

To find the mean, start by adding all of the numbers except for the outliers.

19 + 19 + 19 + 18 + 18 + 18 + 18 + 20 = 149

We added 8 numbers, so divide the sum by 8.

149/8 = 18.6

Therefore, the mean without outliers is 18.6 years old.

Median = middle (order numbers from least to greatest, then find the middle)

To find the median, order the numbers from least to greatest.

18, 18, 18, 18, 19, 19, 19, 20

Now, find the middle.

(Since there are an even number of numbers, select the two numbers in the middle.)

18, 18, 18, 18, 19, 19, 19, 20

Now, add these numbers and divide by 2.

18 + 19 = 37

37/2 = 18.5

Therefore, the median without outliers is 18.5 years old.

Answer:

Step-by-step explanation:

Apply the  priority rule,

(458+56)-134=514-134=380

Answer:

219.91cm²

Step-by-step explanation:

The question is not properly structured. Here is the complete question.

A sector with a radius of 10cm and has a central angle of 252°. Calculate the area of the sector.

Area of a sector = theta/360 × πr²

theta is the angle substended by the circle

r is the radius of the circle.

Given radius = 10cm

theta = 252°

Area of a sector = 252°/360° × π(10)²

Area of a sector = 252/360 × 100π

Area of a sector = 79168.13.360

Answer:

Step-by-step explanation:

The question is not properly structured. Here is the complete question.

A sector with a radius of 10cm and has a central angle of 252°. Calculate the area of the sector.

Area of a sector = theta/360 × πr²

theta is the angle substended by the circle

r is the radius of the circle.

Given radius = 10cm

theta = 252°

Area of a sector = 252°/360° × π(10)²

Area of a sector = 252/360 × 100π

Area of the sector = 219.91cm²

Answer:

70pi

Step-by-step explanation:

That is the answer on khan academy

Answer:

a) In the step-by-step explanation.

b)  The null and alternative hypothesis are:

[tex]H_0: \mu=125000\\\\H_a:\mu> 125000[/tex]

c) P-value = 0.015

d) The conclusion, at a level of significance of 0.05, is that there is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.

e) The claim of the chamber of commerce is false. We have statistical evidence against that claim and we can conclude that the mean prices are significantly higher than $125,000.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim, expressed in the alternative hypothesis, is that area residential property is available at a mean cost significantly higher than $125,000 per lot.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=125000\\\\H_a:\mu> 125000[/tex]

The significance level is 0.05.

The sample has a size n=32.

The sample mean is M=130000.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12500.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12500}{\sqrt{32}}=2209.71[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{130000-125000}{2209.71}=\dfrac{5000}{2209.71}=2.26[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=32-1=31[/tex]

This test is a right-tailed test, with 31 degrees of freedom and t=2.26, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t>2.26)=0.015[/tex]

As the P-value (0.015) 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 that area residential property is available at a mean cost significantly higher than $125,000 per lot.

Answer:

C. 7, 14, 21, 28

Step-by-step explanation:

We want to find the multiples of 7

7*1= 7

7*2 = 14

7*3= 21

7*4 = 28

Answer:

The answer is, C. 7, 14, 21, 28

Answer:

X=70 :) Good Luck!!

Step-by-step explanation:

Using the concept of supplementary angles, it is found that the value of x is of 70 degrees.

-----------------

Thus:

[tex]45 + 2x - 5 = 180[/tex]

[tex]2x + 40 = 180[/tex]

[tex]2x = 140[/tex]

[tex]x = \frac{140}{2}[/tex]

[tex]x = 70[/tex]

The value of x is of 70 degrees.

A similar problem is given at https://brainly.com/question/22826236

Answer:

Step-by-step explanation:

a)

Class limits       Frequency

38 - 63                  9

64 - 89                  7

90 - 115                 1

116 - 141                 1

142 - 167              2

Construct a histogram of these data(image attached)

b)

Frequency distribution table (also known as frequency table) consists of various components.

Classes: A large number of observations varying in a wide range are usually classified in several groups according to the size of their values. Each of these groups is defined by an interval called class interval. The class interval between 10 and 20 is defined as 10-20.

Class limits: The smallest and largest possible values in each class of a frequency distribution table are known as class limits. For the class 10-20, the class limits are 10 and 20. 10 is called the lower class limit and 20 is called the upper class limit.

Class limit: Class limit is the midmost value of the class interval. It is also known as the mid value.

Mid value of each class = (lower limit + Upper limit)2.  

If the class is 38 to 63 , lower limit is 38 and upper limit is 63. So the mid value is

[tex]\frac{(38+63)}{2} =50.5[/tex]

Class width is calculated using upper and lower value

Class width = (upper - lower) value /class

[tex]=\frac{ ( 163 - 38)}{5} \\\\=25[/tex]

Magnitude of a class interval: The difference between the upper and lower limit of a class is called the magnitude of a class interval.

Class frequency: The number of observation falling within a class interval is called class frequency of that class interval.

Answer:

The midpoint of (-2,1) is -1. The midpoint for (-1,1) is 0.

Step-by-step explanation:

To find the midpoint you will have to add x+y then divide by 2.

Answer:

A =754 ft^2

Step-by-step explanation:

The area of a rectangle is given by

A = l*w

A = 58*13

A =754 ft^2

Answer:

C)

Step-by-step explanation:

3x+2>5

3x>3

x>1

I believe that you mistyped C so I think its C

Answer:

Tina can select and answer the essay questions on her test in 35 ways.

Step-by-step explanation:

The order in which she chooses the questions is not important. For example, choosing questions 1, 3 and 7 is the same as choosing 7, 3 and 1. 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:

Three questions from a set of 7. So

[tex]C_{7,3} = \frac{7!}{3!(7-4)!} = 35[/tex]

Tina can select and answer the essay questions on her test in 35 ways.

Tina can select and answer the essay question on her test in 35 ways.

     Number of ways to select 'r' objects out of 'n' objects is given by the expression,

Number of ways = [tex]^nC_r[/tex]

Since, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

If n = 7 and r = 3,

Number of ways to select 3 essays out of 7 will be,

Number of ways = [tex]^7C_3[/tex]

                           [tex]=\frac{7!}{3!(7-3)!}[/tex]

                           [tex]=\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}[/tex]

                           [tex]=35[/tex]

  Therefore, Tina can select and answer the essay question on her test in 35 ways.

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Answer:

The 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40). This means that we are 95% sure that the true proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is between (0.34 and 0.40).

The lower bound of the confidene interval is higher than 0.33. So it is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football, this estimate should be higher than 33%.

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 = 1000, \pi = 0.37[/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}} = 0.37 - 1.96\sqrt{\frac{0.37*0.63}{1000}} = 0.34[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 + 1.96\sqrt{\frac{0.37*0.63}{1000}} = 0.40[/tex]

The 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40). This means that we are 95% sure that the true proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is between (0.34 and 0.40).

Based on your answer to the previous problem, is it reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

The lower bound of the confidene interval is higher than 0.33. So it is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football, this estimate should be higher than 33%.

The 95% Confidence Interval we will found for given case is:

(0.34, 0.40)

It is Not reasonable to believe that  33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.

When there is n sample size and the probability of success is p with level of significance [tex]\alpha[/tex]calculated as:

[tex]CI = p \pm z_0 \sqrt{\dfrac{p(1-p)}{n}[/tex]

Where [tex]z_0[/tex] is the z score  [tex]1- \dfrac{\alpha}{2}[/tex] .

1 -  [tex]\alpha[/tex] = 0.95 ( since it is given to find 95% CI)

Thus,  [tex]\alpha[/tex]  = 0.05

The z score at p value = 1 -  [tex]\alpha[/tex]/2 = 1 -0.025 = 0.975 is 1.96

Thus we have:

[tex]CI = p \pm z_0 \sqrt{\dfrac{p(1-p)}{n}}\\\\CI = 0.37 \pm 1.96 \sqrt{\dfrac{0.37 \times0.63 }{1000}}\\\\CI = 0.37 \pm 1.96 \times 0.0152 = 0.37 \pm 0.03\\[/tex]

Lower limit of CI = 0.34 and Upper limit of CI = 0.40.

The case to decide to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football can be solved as:

33% = 0.33 probability of actually liking watching american probability.

Since 0.33 is lower than the lower limit of CI for 0.37 probability, thus we reject the null hypothesis that both 0.37 and 0.33 are approximately same, and thus decide that it is false that 0.33 percent believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football.

Thus it is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football.

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