Whats the answer to the problem in the picture

The width of a rectangle is seven less than one-fifth its length. If L is used to represent the length of the rectangle, then ho

Dima020 [189]

Answer:check the pic

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

PLEASE HELP FAST!!! AND EXPLAIN YOUR WORK

fomenos

The drummers final position would be
-1, -2

8 0

2 years ago

An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry ou

Katyanochek1 [597]

Answer:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

$p_v =P(t_{(5)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation

The mean and sample deviation can be calculated from the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$

$\bar X=29.35$ represent the sample mean

$s=1.365$ represent the sample standard deviation

$n=6$ sample size

$\mu_o =30$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :

Null hypothesis: $\mu \geq 30$

Alternative hypothesis: $\mu < 30$

Since we don't know the population deviation, 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:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=6-1=5$

Since is a one-side left tailed test the p value would given by:

$p_v =P(t_{(5)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

5 0

2 years ago

7. A man makes a simple discount note for $6,200, at an ordinary bank discount rate of 8.84%, for 40 days. What is the effective

Elina [12.6K]

Answer:

The effective interest rate, rounded to the nearest tenth, is 0.1%.

Step-by-step explanation:

The banker's rule is the simple interest formula.

The simple interest formula is given by:

$E = P*I*t$

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly) and t is the time, in years.

The effective interest rate is given by the following formula:

$E_{IR} = \frac{E}{P}$ .

In this problem, we have that:

A man makes a simple discount note for $6,200, at an ordinary bank discount rate of 8.84%, for 40 days. We consider that the year has 360 days. This means that $P = 6200, I = 0.0884, t = \frac{40}{360} = \frac{1}{9}$ .