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Montano1993 [528]
3 years ago
13

I need help with #39-42, please explain.

Mathematics
1 answer:
shusha [124]3 years ago
3 0

For question 39 & 40, we need to use the below equation to complete the sentence

l=m\sqrt{n}

Question 39:

When ' n ' increase, the \sqrt{n} will also increase and that multiplied with constant ' m ', the l will also increase.

Solution for question 39:

As n increases and m stays constant , l <u>increases</u>

-------

Question 40:

Solving the equation for m, we get

l = m\sqrt{n} \\ \\ m=\frac{l}{\sqrt{n}}

When ' l ' increases, the numerator increase, the denominator stays constant because 'n' stays constant, for this condition, the fraction increases.

Solution for question 40:

As l increase and n stays constant, m <em><u>increases</u></em>

------

For question 41 & 42, we need to use the below equation to complete the sentence

r=s^2/t^2

Question 41:

When s is triped, the equation will be...

r=(3s)^2/t^2=\frac{3^{2}s^{2}}{t^2}   =9s^2/t^2

Solution for question 41:

If s is tripled and t stays constant, r is multiplied by <em><u>9</u></em>

--------

Question 42:

When t is doubled, the equation will be...

r=s^2/(2t)^2=\frac{s^2}{2^2 \cdot t^2}=\frac{s^2}{4t^2}\\   \\ r=0.25s^2/t^2 \; \; (or) \; \; \frac{1}{4} \cdot \frac{s^2}{t^2}

Solution for 42:

If t doubled and s stays constant, r is multiplied by <em><u>1/4 or 0.25</u></em>


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Step-by-step explanation:

Let's look at choice 1.

x=t+1

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I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

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Let's look at choice 2.

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y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

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y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

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Answer:

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

df=n-1=28-1=27  

p_v =P(t_{(27)}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=60 represent the sample mean

s=8.9 represent the sample standard deviation

n=28 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 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 (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 63, the system of hypothesis would be:  

Null hypothesis:\mu \geq 63  

Alternative hypothesis:\mu < 63  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=28-1=27  

Since is a one side lower test the p value would be:  

p_v =P(t_{(27)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

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