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3 years ago

Shana wants to use all 62 feet of the fencing she has to make a rectangular run for her dog. She decides to make the

1 answer:

5 0

Answer:The value of w cannot be a negative number, substitution is used to replace the variable L with a value of 20, the subtraction property of equality is used to isolate the term with the variable w.

Step-by-step explanation:

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In the image, the distance from the swings to the slide is 35

Nikitich [7]

The Pythagorean theorem is a^2+b^2=c^2 where a and b are legs and c is the hypotenuse(longest side). you haven’t put a picture so I don’t know where 35 is a leg or the hypotenuse, but all you have to do is square the 2 shorter sides and add them and set it equal to the square of the longer side. You can do it!

5 0

3 years ago

A student claims that 8^3 x 8^-5 is greater than 1. Explain whether the student is correct or not.

nlexa [21]

Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m

8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1

Answer: the student is not correct, because 1/64<1, and 8³ x 8⁻⁵=1/64.

7 0

3 years ago

the case of napkins contained 45 packs with a total of 10,125 napkins. each napkin contained the same number of napkins. write a

Pavlova-9 [17]

You would need to do 10,125/45 = n, and in which case we just solve.
10,215/45=227, so there are 227 napkins in each box.

Hope this helps and have a nice day:)

7 0

3 years ago

Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.50ft.

Anarel [89]

Dimensions of the room in cm = 2.54 x 12 by 15 x 2.54 by 2.54 x 8.5 = 30.48 by 38.1 by 21.59

Volume of the room in cubic cm = 30.48 x 38.1 x 21.59 cubic cm = 25,072.21 cubic cm

Given that the density of air at room temperature is $0.00118g/cm^3$ , thus the mass of air in the room = 25,072.21 x 0.00118 = 29.59 g = 0.0296 kg

Given that the lethal dose of HCN is approximately 300 mg HCN per kilogram of air when inhaled, thus the <span>amount of HCN that gives the lethal dose in the small laboratory room is given by 300 x 0.0296 = 8.88 mg.</span>

7 0

3 years ago

A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life

alisha [4.7K]

Answer:

The null hypothesis is $H_0: \mu \leq 10$

The alternative hypothesis is $H_1: \mu > 10$

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

$H_0: \mu \leq 10$

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

$H_1: \mu > 10$

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

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

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that $\mu = 10$

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that $n = 45, X = 10.5, s = 1.8$

Then

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

$t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}$

$t = 1.86$

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

5 0

2 years ago