Tom bought n DVDs for a total cost of 15n - 2 dollars. Which expression represents the cost of each DVD?

melamori03 [73]

Answer: ar·ith·met·ic pro·gres·sion

/ˌeriTHˌmedik prəˈɡreSH(ə)n/

noun

a sequence of numbers in which each differs from the preceding by a constant quantity (e.g., 3, 6, 9, 12, etc.; 9, 7, 5, 3, etc.).

the relationship between numbers in an arithmetic progression.

"the numbers are in arithmetic progression"

Step-by-step explanation:

4 0

2 years ago

Write parametric equations of the line 2x-y=3

dangina [55]

In this item we are given with the equation, 2x - y = 3. The equation contains two variables, x and y. We assume in this item that the value of x is independent of the value of y; however, y values depends on the given values of x. In parametric form, the equation would take the form,

f(x) = y = ax + b

where a is the numerical coefficient of x and b is constant. Transforming the given equation to this form,

f(x) = y = 2x - 3

4 0

3 years ago

Read 2 more answers

The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.

valina [46]

Answer:

A sample of 179 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.85}{2} = 0.075$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.075 = 0.925$ , so Z = 1.44.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study found that for an average family the variance is 1.69 gallon?

This means that $\sigma = \sqrt{1.69} = 1.3$

If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?

A sample of n is needed, and n is found for M = 0.14. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.14 = 1.44\frac{1.3}{\sqrt{n}}$

$0.14\sqrt{n} = 1.44*1.3$

$\sqrt{n} = \frac{1.44*1.3}{0.14}$

$(\sqrt{n})^2 = (\frac{1.44*1.3}{0.14})^2$

$n = 178.8$

Rounding up

A sample of 179 is needed.

7 0

3 years ago

2564.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A

Leokris [45]

Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let $\mu$ = population mean length of the board.

So, Null Hypothesis, $H_0$ : $\mu$ = 2564.0 millimeters {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 2564.0 millimeters {means that the boards are either too long or too short}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s }{\sqrt{n}} }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean length of boards = 2559.5 millimeters

s = sample standard deviation = 15.0 millimeters

n = sample of boards = 26

So, the test statistics = $\frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }$ ~ $t_2_5$

= -1.529

The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

4 0

4 years ago

What is the value of the expression?

nadya68 [22]

B use the app Photomath it helps you with this kinda questions

8 0

3 years ago