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Lana71 [14]
3 years ago
9

A company can sell 2000 magazine subscriptions at $40 each. For each $5 increase in the price, it will sell 200 fewer subscripti

ons. What subscription price will provide the maximum revenue for the company?
$55

$40

$45

$50
Mathematics
1 answer:
Olenka [21]3 years ago
4 0

<u>ANSWER</u>

The subscription price of $45 provides the maximum revenue for the company


<u>EXPLANATION</u>

When the price is $40, the company sells 2000 quantities.

The total revenue accrued

=40\times2000


=$80,000


When the price increases to $45 the quantity sold will decrease by 200 magazines which is 2000-200=1800


The total revenue accrued

=45\times 1800


=81000 dollars


When the price increases to $50 the quantity sold will decrease by 200 magazines again which is 1800-200=1600


The total revenue accrued

=50\times 1600


=80000 dollars


When the price increases to $55 the quantity sold will decrease by 200 magazines which is 1600-200=1400


The total revenue accrued

=55\times 1400


=77000 dollars


Hence a subscription price of $45 provides the maximum revenue for the company.



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Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes n_{1} = 400 and n_{2} = 300

Proportion of mean p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52

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<u>Alternative hypothesis H1:</u>- p1 ≠ p2

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Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }

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Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }\\   =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300}   }

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

<u>conclusion</u>:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) <u>95% confidence intervals</u>

The confidence intervals are P± 1.96(√PQ/n)

we know that = p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}

after calculation we get P = 0.56 and Q =1-P =0.44

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on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

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