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MrRissso [65]
3 years ago
9

You are trying to decide if you want to buy a seasonal golf membership for $224. If you buy the membership, you can golf as many

games as you want during the season and can rent golf clubs for free. Without the pass, it will cost you $24 per game plus an additional $4 to rent golf clubs each game. If you need to ren clubs, after how many games of golf will the cost of the membership be less than the cost of golfing without the membership?
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
6 0

Answer:

Nine games

Step-by-step explanation:

The cost of a membership is $224. The cost of a golf game and renting golf clubs is $28. The cost of nine games of golf and renting golf clubs is $252  (28 x 9= 252), so it is more than $224.

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x = 64/4 = 16°.

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2x-3y = 116°

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The weight of bags of coffee beans at the grocery store follows a Normal distribution with a mean of μ = 7 ounces and a standard
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Answer:

C.

Step-by-step explanation:

The random variable W is Normal, with a mean of 28 ounces and a standard deviation of four ounces.  

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Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampl
galben [10]

The missing part of the question is highlighted in bold format

The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age at first startup for 10% of entrepreneurs was 30 years of age or more.

Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. If required, round your answers to four decimal places. np = n(1-p) = E(p) = σ(p) = (b) Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. If required, round your answers to four decimal places.

Answer:

(a)

np = 180

n(1-p) = 20

E(p) = p = 0.9

σ(p) = 0.0212

(b)

np =  20

n(1 - p) = 180

E(p) = p = 0.1

σ(p) = 0.0212

Step-by-step explanation:

From the given information:

Let consider p to be the sample proportion of entrepreneurs whose first startup was at 29 years of age or less

So;

Given that :

p = 90%    i.e  p = 0.9

sample size n = 200

Then;

np = 200 × 0.9 = 180

n(1-p) = 200 ( 1 - 0.9)

= 200 (0.1)

= 20

Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;

Then:

The expected value of the sampling distribution of p =  E(p) = p = 0.9

Variance \sigma^2=\dfrac{p(1-p)}{n}

=\dfrac{0.9(1-0.9)}{200}

=\dfrac{0.9(0.1)}{200}

\mathbf{=4.5*10^{-4}}

The standard error of  σ(p) = \sqrt{\sigma ^2}

\mathbf{= \sqrt{4.5*10^{-4}}}

= 0.0212

(b)

Here ;

p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more

p = 10%   i.e p = 0.1

sample size n = 200

Then;

np = 200 × 0.1 = 20

n(1 - p) = 200 (1 - 0.1 )  = 180

Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;

Then:

The Expected value of the sampling distribution of p = E(p) = p = 0.1

Variance \sigma^2=\dfrac{p(1-p)}{n}

=\dfrac{0.1(1-0.1)}{200}

=\dfrac{0.1(0.9)}{200}

\mathbf{=4.5*10^{-4}}

The standard error of  σ(p) = \sqrt{\sigma ^2}

\mathbf{= \sqrt{4.5*10^{-4}}}

= 0.0212

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Answer:

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Next time, please try to make your question more specific.

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