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AleksandrR [38]
3 years ago
11

A study found that a​ student's GPA,​ g, is related to the number of hours worked each​ week, h, by the equation g= −0.0006h^2 +

0.016h +3.02 Estimate the number of hours worked each week for a student with a GPA of 2.23
Mathematics
2 answers:
elena55 [62]3 years ago
6 0

Answer:

51.9914 or 52 hrs.

Step-by-step explanation:

-0.0006h²+0.016+0.79 (divide each term by -0.0006)

=h²-(80/3)h-(3950/3)

Using the quadratic formula

((80/3)+sqrt((-80/3)^2-4(1)(-3950/3)))/2(1)

= 103.9828/2

=52.9914

Aprroximation: 51.9914 or 52 hrs.

Exact: ((80/3)+sqrt(53800/3))/2

faltersainse [42]3 years ago
6 0

Answer:

The number of hours worked each week for a student with a GPA of 2.23 is 50 hours.      

Step-by-step explanation:

Given : A study found that a​ student's GPA,​ g, is related to the number of hours worked each​ week, h, by the equation  g= -0.0006h^2 +0.016h +3.02

To find : Estimate the number of hours worked each week for a student with a GPA of 2.23 ?

Solution :

The equation representing situation is

g= -0.0006h^2 +0.016h +3.02

Now, Substitute g=2.23

2.23= -0.0006h^2 +0.016h +3.02

Solve,

0.0006h^2-0.016h-0.79=0

Multiply both side by 10000,

6h^2-160h-7900=0

Apply quadratic formula,

h=\frac{-(-160)\pm\sqrt{(-160)6-4(6)(7900)}}{2(6)}

h=\frac{160\pm\sqrt{188640}}{12}

h=\frac{160+\sqrt{188640}}{12},\frac{160-\sqrt{188640}}{12}

h=49.52,-22.8

Reject h=-22.8

Approximately the value of h is 50.

Therefore, The number of hours worked each week for a student with a GPA of 2.23 is 50 hours.

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Step-by-step explanation:

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Of the total population of American households, including older Americans and perhaps some not so old, 17.3% receive retirement
Alex Ar [27]

Answer:

47.54% probability that more than 20 households but fewer than 35 households receive a retirement income

Step-by-step explanation:

We use the binomial aproxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.173, n = 120. So

\mu = E(X) = np = 120*0.173 = 20.76

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.173*0.827} = 4.14

In a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?

We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.

X = 34

Z = \frac{X - \mu}{\sigma}

Z = \frac{34 - 20.76}{4.14}

Z = 3.2

Z = 3.2 has a pvalue of 0.9993

X = 21

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 20.76}{4.14}

Z = 0.06

Z = 0.06 has a pvalue of 0.5239

0.9993 - 0.5239 = 0.4754

47.54% probability that more than 20 households but fewer than 35 households receive a retirement income

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4 years ago
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