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sertanlavr [38]
3 years ago
5

The Kamp family has twins, Rob and Rachel. Both Rob and Rachel graduated from college 2 years ago, and each is now earning $50,0

00 per year. Rob is an engineer. The mean salary for engineers with less than 5 years’ experience is $60,000 with a standard deviation of $5,000. Rachel works in the retail industry, where the mean salary for executives with less than 5 years’ experience is $35,000 with a standard deviation of $8,000.
Compute the z values for both Rob and Rachel and comment on your findings.
Mathematics
1 answer:
dybincka [34]3 years ago
5 0

Answer:

z-value of rachel = 1.875

z-value of rob = -2

z-value of Rachel is more than that of rob. Thus rob is earning below average and rachel is earning above average.

Step-by-step explanation:

Let's denote the salary of Rob and Rachel per year by X. So, X = $50,000

We are told that;

For Rachel's industry;

Mean salary;μ1 = $35,000

Standard deviation;σ1 = $8,000

For Rob's industry;

Mean salary;μ2 = $60,000

Standard deviation;σ2 = $5,000

Formula for z - value is;

z = (X - μ)/σ

Thus;

z-value for rob is;

z2 = (X - μ2)/σ2

z2 = (50000 - 60000)/5000

z2 = -2

z-value for rachel is;

z1 = (X - μ1)/σ1

z1 = (50000 - 35000)/8000

z1 = 1.875

z-value of Rachel is more than that of rob. Thus rob is earning below average and rachel is earning above average.

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<span><u><em>Answer:</em></u>
(5 * 1) + (9 * 0.1).<span>

<u><em>Explanation:</em></u>
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The positions when the particle reverses direction are:

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Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

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We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

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v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

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v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

We know that the particle reversed its direction two times.

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a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42

Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

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