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Anika [276]
3 years ago
10

A tire company is selling two different tread patterns of tires. Tire x sells for $75.00 and tire y sells for $85.00.Three times

the number of tire y sold must be less than or equal to twice the number of x tires sold. The company has at most 300 tires to sell.
What is the maximum revenue that the company can make?



$13,500

$22,500

$23,700

$25,500

Mathematics
2 answers:
bija089 [108]3 years ago
4 0

Let

x-------> the number of x tires sold

y-------> the number of y tires sold

we know that

3y \leq 2x ------> equation 1

x+y \leq 300 ------> equation 2

using a graph tool

see the attached figure

the solution is the shaded area

the maximum revenue that the company can make is for the point (180,120)

x=180\ tires\\ y=120\ tires\\ Revenue=75*180+85*120\\ Revenue=23,700

therefore

the answer is the option

$23,700

kirza4 [7]3 years ago
3 0
23,700 is the revenue they can make at most
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Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

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Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

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       σ/√n

The formula for the interval is:

X[bar] ± Z_{1- \alpha /2} * (σ/√n)

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a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

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2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

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"point estamator" ± "margin of error"

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d= Z_{1- \alpha /2} * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= Z_{1- \alpha /2} * (σ/√n)

\frac{d}{Z_{1- \alpha /2}}= σ/√n

√n * \frac{d}{Z_{1- \alpha /2}}= σ

√n = σ * \frac{Z_1- \alpha /2}{d}

n = (σ * \frac{Z_1- \alpha /2}{d})²

n=  (0.33 * \frac{1.28}{0.08})²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

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