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Julli [10]
4 years ago
6

A fruit grower knows from previous experience and careful data analysis that if the fruit on a specific kind of tree is harveste

d at this time of year, each tree will yield, on average, 137 pounds, and will sell for $2.4 per pound. However, for each additional week the harvest is delayed (up to a point), the yield per tree will increase by 2.5 pounds, while the price per pound will decrease by $0.04.
Mathematics
1 answer:
Lostsunrise [7]4 years ago
4 0

Answer:

A) t = 2.6 weeks

B) R = $329.48

Step-by-step explanation:

At the right time, the average amount of money made from the fruits is given as

R = (selling price of fruits per pound) × (average number of pounds of fruits) = 2.4 × 137 = $328.8 on average.

So, delaying the harvest by a week makes the revenue turn to

R = (2.4 - 0.04)(137 + 2.5)

Delaying the harvest by t weeks makes the revenue turn to

R = (2.4 - 0.04t) × (137 + 2.5t)

R = 328.8 + 6t - 5.48t - 0.1t²

R(t) = -0.1t² + 0.52t + 328.8

A) Time in weeks when the Revenue would be highest.

That is, value of t that corresponds to the maximum value of R(t)

At maximum value, (dR/dt) = 0 and (d²R/dt²) < 0, that is negative.

R(t) = -0.1t² + 0.52t + 328.8

(dR/dt) = -0.2t + 0.52 = 0

and (d²R/dt²) = -0.2 < 0 (negative; this is indeed a maximum point for the function)

When (dR/dt) = 0,

(dR/dt) = -0.2t + 0.52 = 0

-0.2t + 0.52 = 0

t = (0.52/0.2)

t = 2.6 weeks.

B) Actual Maximum value of the Revenue.

This corresponds to the revenue when t = 2.6 weeks

R(t) = -0.1t² + 0.52t + 328.8

R(2.6) = -0.1(2.6)² + 0.52(2.6) + 328.8

R(t = 2.6) = -0.676 + 1.352 + 328.8

R at maximum point = $329.48

Hope this Helps!!!

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Answer:

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The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

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For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=46 represent the mean height for the sample  

s=5.148 represent the sample standard deviation

n=5 sample size  

\mu_o =40 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:  

Null hypothesis:\mu = 40  

Alternative hypothesis:\mu \neq 40  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing we got:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

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Use that in (1)

x+2*(-1/2)=0

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