MATHEMATICS COLLEGE

Given the exponential equation y = yoxQq describe each of the following below:y is
yo is
x is
Q is
q is

Answers

Answer 1
Answer:

Answer:

Step-by-step explanation:

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Jan raised $120.75 at the car wash. If each wash costs $5.75, how many cars did she wash?

Answers

Answer:

She washed 21 cars

Step-by-step explanation:

Answer:

Jan washed 21 cars

Step-by-step explanation:

120.75 / (Divided) 5.75=21

If you were to times 5.75 by 21 you will get the amount she rasied at the car wash

Assume that the standard deviation of daily returns for Marcus, Inc. stock in a recent period is 1.5 percent. Furthermore, a 95 percent confidence interval is desired for the maximum loss. Daily returns are normally distributed, and the expected daily return is 0.05 percent. What is the lower boundary of the maximum expected loss

Answers

Answer:

= - 2.43%

Step-by-step explanation:

From the information given:

Since the variable (daily returns) is normally distributed, Then, using empirical rule at 95% confidence interval level, we have:

( \mu - 1.96 \sigma \ , \ \mu + 1.96 \sigma)

where;

The expected mean daily return \mu = 0.05 \%

The standard deviation \sigma = 1.5\%

Given that the 95% confidence interval is expected to be a maximum loss, then the probability is left-tailed which is 1.65\sigma away from the average.

Thus the distribution of the lower boundary can be computed as:

= (0.05 - 1.65 * 1.5)\%

= (0.05 - 2.475)\%

= ( - 2.425)\%

= - 2.43%

Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute indicate that the percentage is now 46% (the Wall Street Journal, October 5, 2012)a. Develop appropriate hypotheses such that rejection of H0 will support the conclusion that a smaller proportion of American families own stocks or stock funds in 2012 than 10 years ago.
b. Assume the Investment Company Institute sampled 300 American families to estimate that the percent owning stocks or stock funds was 46% in 2012. What is the p-value for your hypothesis test?
c. At α = .01, what is your conclusion?

Answers

Using the z-distribution, as we are working with a proportion, it is found that:

a) H_0: p = 0.53, H_1: p < 0.53

b) The p-value is of 0.0075.

c) Since the p-value of the test is of 0.0075 < 0.01 for the left-tailed test, it is found that there is enough evidence to reject the null hypothesis and conclude that a smaller proportion of American families own stocks or stock funds in 2012 than 10 years ago.

What are the hypothesis tested?

At the null hypothesis, it is tested if the proportion is still of 53%, that is:

H_0: p = 0.53

At the alternative hypothesis, it is tested if the proportion is now smaller, that is:

H_1: p < 0.53

Item a:

The hypothesis are:

H_0: p = 0.53

H_1: p < 0.53

Item b:

The test statistic is given by:

z = \frac{\overline{p} - p}{\sqrt{(p(1-p))/(n)}}

In which:

  • \overline{p} is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

In this problem, the parameters are:

\overline{p} = 0.46, p = 0.53, n = 300.

Hence, the value of the test statistic is given by:

z = \frac{\overline{p} - p}{\sqrt{(p(1-p))/(n)}}

z = \frac{0.46 - 0.53}{\sqrt{(0.53(0.47))/(300)}}

z = -2.43

Using a z-distribution calculator, considering a left-tailed test, as we are testing if the proportion is less than a value, with z = -2.43, it is found that the p-value is of 0.0075.

Item c:

Since the p-value of the test is of 0.0075 < 0.01 for the left-tailed test, it is found that there is enough evidence to reject the null hypothesis and conclude that a smaller proportion of American families own stocks or stock funds in 2012 than 10 years ago.

More can be learned about the z-distribution at brainly.com/question/26454209

Answer:

a) Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

b) z=\frac{0.46 -0.53}{\sqrt{(0.53(1-0.53))/(300)}}=-2.429  

p_v =P(Z<-2.429)=0.0076

c) So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .  

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=0.46 estimated proportion of American families owning stocks or stock funds

p_o=0.53 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:  

Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.46 -0.53}{\sqrt{(0.53(1-0.53))/(300)}}=-2.429  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z<-2.429)=0.0076

Part c  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .  

2 to the Third power -(30÷5) +9

Answers

Answer:

11

Step-by-step explanation:

2^3-(30/5)+9

First, you divide 30/5 because it is in parentheses and we are following this rule for order of operations:

P- parentheses

E- exponents

M- multiplication

D- division

A- addition

S- subtraction

so 30/5 = 6 and plugged in is:

2^3-(6)+9

Next, we do exponents:

2^3 = 2x2x2 = 8

plugged back in:

8-(6)+9

Now, we do the math from left to right so:

8-6=2

2+9=11

11 is your answer

Find the total surface area of this cubiod

Answers

Answer:

148 cm^2.

Step-by-step explanation:

It has 2 surfaces of area 5*6, 2 of  5*4 and 2 of 4*6.

= 2*30 + 2*20 + 2*24

= 60 + 40 + 48

= 148 cm^2.

Find the future value (FV) of the annuity due. (Round your answer to the nearest cent.) $175 monthly payment, 7% interest, 11 years

Answers

Answer:

The future value of the annuity due to the nearest cent is $2956.

Step-by-step explanation:

Consider the provided information:

It is provided that monthly payment is $175, interest is 7% and time is 11 years.

The formula for the future value of the annuity due is:

FV of Annuity Due = (1+r)* P[((1+r)^(n)-1)/(r)]

Now, substitute P = 175, r = 0.07 and t = 11 in above formula.

FV of Annuity Due = (1+0.07)* 175[((1+0.07)^(11)-1)/(0.07)]

FV of Annuity Due = (1.07)* 175[(1.10485)/(0.07)]

FV of Annuity Due = 187.5(15.7835)

FV of Annuity Due = 2955.4789

Hence, the future value of the annuity due to the nearest cent is $2956.
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