What is the effective annual yield. 360 days in a year. 9.5%compounded monthly.

Elimination Method 7x=3y+45 4x+5y=19 (Show it step by step please!)

Fittoniya [83]

$\left\{\begin{array}{ccc}7x=3y+45&|\text{subtract 3y from both sides}\\4x+5y=19\end{array}\right\\\left\{\begin{array}{ccc}7x-3y=45&|\text{multiply both sides by 5}\\4x+5y=19&|\text{multiply both sides by 3}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}35x-15y=225\\12x+15y=57\end{array}\right}\qquad|\text{add both sides of the equations}\\.\qquad47x=282\qquad\text{divide both sides by 47}\\.\qquad \boxed{x=6}\\\\\text{Put the value of x to the second equation}$

$4(6)+5y=19\\\\24+5y=19\qquad\text{subtract 24 from both sides}\\\\5y=-5\qquad\text{divide both sides by 5}\\\\\boxed{y=-1}\\\\Answer:\ \boxed{x=6\ and\ y=-1}$

8 0

4 years ago

In a multiple linear regression analysis, statistical model utility is determined by the value of R2 and 2s, while practical mod

Natali [406]

Answer:

This statement is true because if F statistics is significant, then the entire multiple regression model is useful for the prediction of y.

Step-by-step explanation:

Solution:

In a multiple linear regression analysis, the statistical model utility is determined by the value of R2 and ∂2, while the Global F test p-value determines the practical model.

This statement is true because if F statistics is significant, then the entire multiple regression model is useful for the prediction of y.

The F test is a statistical test that is very flexible. F test can evaluate multiple model terms.

The F test indicates whether a linear regression model provides a better fit to data than a model that contains no independent variable.

The F test for overall significance has two hypotheses:

1- Null hypothesis:

The fit of the intercept only model and your model are equal.

2- Alternative hypothesis:

The fit of the intercept-only model is significantly reduced compared to your model.

Compare p-value for the F test,

If the p-value is less than the significance level, the sample data provide sufficient evidence that the regression model fits the data better with no independent variable.

5 0

3 years ago

What is the lower fence of this data set?A) 20.5 B) 18 C) 20 D) 19.5

gulaghasi [49]

A, sorry if it’s wrong

4 0

3 years ago

Identifying functions worksheet

cestrela7 [59]

Answer:

Function
Not a function
Not a function
Function
Function
Not a function

7 0

2 years ago

Read 2 more answers

The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April,

Karolina [17]

Answer:

a)Null hypothesis: $\mu = 24.57$

Alternative hypothesis: $\mu \neq 24.57$

b) $df=n-1=30-1=29$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(29)}$

c) If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 24.57 at 5% of signficance.

d) For this case since we have a two tailed test we need to find two critical values and we need 0.025 of the area on each tail of the t distribution with 29 degrees of freedom. The critical values are given by:

$t_{crit}=\pm 2.045$

And the rejection zone would be given by:

$(-\infty , -2.045) U(2.045,\infty)$

Since pur calculated value is not on the rejection zone we Fail to reject the null hypothesis.

Step-by-step explanation:

The mean hourly wage for employees in goods-producing industries is currently $24.57. Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.

a. State the null and alternative hypotheses we should use to test whether the population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the goods-producing industries.

We need to conduct a hypothesis in order to check if the mean is equal to 24.57 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 24.57$

Alternative hypothesis: $\mu \neq 24.57$

b. Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour and compute the p-value.

$\bar X=23.89$ represent the sample mean

$s=2.4$ represent the sample standard deviation

$n=30$ sample size

$\mu_o =24.57$ represent the value that we want to test

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

If we analyze the size for the sample is = 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{23.89-24.57}{\frac{2.4}{\sqrt{30}}}=-1.552$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=30-1=29$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(29)}$

c. With α = 0.05 as the level of significance, what is your conclusion?

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 24.57 at 5% of signficance.

d. Repeat the preceding hypothesis test using the critical value approach.

For this case since we have a two tailed test we need to find two critical values and we need 0.025 of the area on each tail of the t distribution with 29 degrees of freedom. The critical values are given by:

$t_{crit}=\pm 2.045$

And the rejection zone would be given by:

$(-\infty , -2.045) U(2.045,\infty)$

Since pur calculated value is not on the rejection zone we Fail to reject the null hypothesis.

4 0

3 years ago