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JulsSmile [24]
2 years ago
9

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha

rging hot water into the river is given by
T(x) = 160-0.05x^2
Find the average temperature over each interval.
a. [0, 10]
b. [10, 40]
c. [0, 40]
Mathematics
1 answer:
mr Goodwill [35]2 years ago
6 0

Answer:

Step-by-step explanation:

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by

T(x) = 160-0.05x^2

a. [0, 10]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

The average temperature

= (160 + 155)/2 = 157.5

b. [10, 40]

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (80 + 155)/2 = 117.5

c. [0, 40]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

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The theater make the maximum profit at d = $25. Then the maximum profit of the theatre is $ 900.

<h3>What is differentiation?</h3>

The rate of change of a function with respect to the variable is called differentiation. It may be increasing or decreasing.

A community theater uses the function P(d) = (− 4d + 40) (d − 40) to model the profit (in dollars) expected on a weekend when the tickets to a comedy show are priced at d dollars each.

Then the maximum profit of the theatre will be

The function is P(d) = (− 4d + 40) (d − 40)

Differentiate the function with respect to d and put it equal to zero for maximum or minimum profit.

\begin{aligned} \dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= 0\\\\\dfrac{\mathrm{d} }{\mathrm{d} d}(- 4d + 40) (d - 40) &= 0\\\\(-4d+40) -4 (d-40) &= 0\\\\-8d + 200 &= 0\\\\d &= 25 \end{aligned}

Then the checking for maximum or minimum, again differentiate, we have

\begin{aligned} \dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= \dfrac{\mathrm{d} }{\mathrm{d} d}(- 4d + 40) (d - 40) \\\\\dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= \dfrac{\mathrm{d} }{\mathrm{d} d}(-8d + 200) \\\\\dfrac{\mathrm{d} }{\mathrm{d} d}P(d) &= -8\\\\ \dfrac{\mathrm{d} }{\mathrm{d} d}P(d) & < 0\end{aligned}

The value is less than zero hence maximum value will occur at d = 25.

Then maximum profit will be

P(d) = (− 4×25 + 40) (25 − 40)

P(d) = (− 100 + 40) (−15)

P(d) = (− 60) (− 15)

P(d) = $ 900

More about the differentiation link is given below.

brainly.com/question/24062595

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I = Adt; where p = $290 is the Amount of the loan, d = 16% = 0.16 is the discount rate and t = 90days = 90/365 = 18/73 is the period.

I = 290 x 0.16 x 18/73 = 11.44
Interest = $11.44

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

9.)

z + 3 > 2/3

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Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the pre
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Answer:

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

\hat p=\frac{46}{115}=0.4 estimated proportion of people that have traded in their old car

p_o=0.48 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.9

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.48.:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

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