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3 years ago

15

A person earns $187.42 before deductions for 3 days work . What is the rate of per hour assuming 8 hours of work per day ?

1 answer:

Bingel [31]3 years ago

6 0

7.809125 dollars per hour

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How do I solve this?? pic attached

lisov135 [29]

Answer:

y=-⅓x+7

Step-by-step explanation:

Since the new line we need to make must be parallel to the old one the number changing x must stay the same. this means that that we will still have -⅓ X. Now that we have part of the line formula, we can try to fit it to the point. Because the X is equal to 9, -1/3*9 is -3. To get it to the correct point, we have to add 7, so that our Y is 4. If we combine the two parts of the equation, we get Y=-⅓x+7

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3 years ago

Four types of energy?

ohaa [14]

<span>Potential Energy is any type of stored energy. It can be chemical, nuclear, gravitational, or mechanical.Kinetic ...GRAVITATIONAL.CHEMICAL.<span>NUCLEAR.
even motion.</span></span>

3 0

3 years ago

Read 2 more answers

Rafael counted a total of 40 white cars and yellow.There were 9 times as many white cars as yellow cars how many white cars did

Naily [24]

Well 9x40 would be the solution, so what is 9x40? That will be your answer.

6 0

3 years ago

Simply. If the solution is not a real number enter not a real number rotate picture answer all 3 please

yuradex [85]

Answer:

13. $\frac{\sqrt[5]{x^4} }{x}$ .

14. $v = \pm3\sqrt{5}$

15. 2.

Step-by-step explanation:

13. $x^{1/5} * x^{-2/5}$

= $x^{1/5 + (-2/5)}$

= $x^{1/5 - 2/5}$

= $x^{-1/5}$

= $\frac{1}{x^{1/5}}$

= $\frac{x^{4/5}}{x^{1/5 + 4/5}}$

= $\frac{x^{4/5}}{x}$

= $\frac{\sqrt[5]{x^4} }{x}$ .

14. $v^2 - 45 = 0$

$v^2 = 45$

$\sqrt{v^2} = \pm\sqrt{45}$

$\sqrt{v^2} = \pm\sqrt{3^2 * 5}$

$v = \pm3\sqrt{5}$ .

15. $\sqrt[3]{2} * \sqrt[3]{4}$

= $\sqrt[3]{2 * 4}$

= $\sqrt[3]{2 * 2 * 2}$

= $\sqrt[3]{2 ^3}$

= 2.

Hope this helps!

7 0

3 years ago

SAT verbal scores are normally distributed with a mean of 433 and a standard deviation of 90. Use the Empirical Rule to determin

laila [671]

34% of the scores lie between 433 and 523.

Solution:

Given data:

Mean (μ) = 433

Standard deviation (σ) = 90

<u>Empirical rule to determine the percent:</u>

(1) About 68% of all the values lie within 1 standard deviation of the mean.

(2) About 95% of all the values lie within 2 standard deviations of the mean.

(3) About 99.7% of all the values lie within 3 standard deviations of the mean.

$$Z(X)=\frac{x-\mu}{\sigma}$

$$Z(433)=\frac{433-\ 433}{90}=0$

$$Z(523)=\frac{523-\ 433}{90}=1$

Z lies between o and 1.

P(433 < x < 523) = P(0 < Z < 1)

μ = 433 and μ + σ = 433 + 90 = 523

Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.

i. e. $((\mu-\sigma) \ \text{to} \ (\mu+\sigma))=68\%$

Here μ to μ + σ = $\frac{68\%}{2} =34\%$

Hence 34% of the scores lie between 433 and 523.

8 0

3 years ago