How many students did Xavier survey?

In your random sample of 25 Chips Ahoy bags, you find a mean of 985 chips per bag, with a standard deviation of 10 chips. Run a

stealth61 [152]

Answer:

The p -value is less than 0.001.

Step-by-step explanation:

Given information:

Sample size = 25 chips

Sample mean = 985

Sample standard deviation = 10

Let as assume that sample is distributed normally.

The formula for test statistic

$t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$

where,

$\overline{x}$ is sample mean

μ is population mean.

s is sample standard deviation.

n is sample size.

The value of test statistic is

$t=\frac{985-1000}{\frac{10}{\sqrt{25}}}$

$t=-7.5$

The p-value is

$p=2\times p(t$

Therefore the p -value is less than 0.001.

6 0

4 years ago

Find the area of the triangle to the nearest tenth.

s344n2d4d5 [400]

Answer:

It C

Step-by-step explanation:

7 0

3 years ago

What number goes in the box in the followng sequence 5 11 23 ? 95 191

Alina [70]

Answer:

47

Step-by-step explanation:

The number that should go after 23 is 47.

The pattern seems to be (n * 2) + 1, where n is the value of the previous number.

We start with the number 5:

5

(5 * 2) + 1 = 11

11

(11 * 2) + 1 = 23

23

(23 * 2) + 1 = 47

47

... and so on

The pattern should continue as follows.

Hope this helps!

3 0

3 years ago

Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.

Maru [420]

Answer:

$9\text{ln}|x|+2\sqrt{x}+x+C$

Step-by-step explanation:

We have been an integral $\int \frac{9+\sqrt{x}+x}{x}dx$ . We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

$\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx$

$\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx$

Now, we will apply the sum rule of integrals as:

$\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx$

$9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx$

Using common integral $\int \frac{1}{x}dx=\text{ln}|x|$ , we will get:

$9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx$

Now, we will use power rule of integrals as:

$9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx$

$9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2\sqrt{x}+\int 1dx$

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

$9\text{ln}|x|+2\sqrt{x}+x+C$

Therefore, our required integral would be $9\text{ln}|x|+2\sqrt{x}+x+C$ .

4 0

3 years ago

Mrs. Ellington has a $50 gift card to Best Buy. She purchased a DVD for $28. Which inequality shows all possible values for x, t

Effectus [21]

Answer:

4th option

Step-by-step explanation:

She purchased a DVD for $28.

$50 - $28 = $22

1. 22 < 22

2. 22 > 22

3. 22 < 78

4. 22 > 78

I would say it would be the 4th option because she can only spend less than 78.

I can be very incorrect though, don't depend on me that much. Just make sure to think about my answer and what you would answer.

4 0

3 years ago

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