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Mariana [72]
3 years ago
14

Suppose f(x) = x^2 and

Mathematics
1 answer:
SpyIntel [72]3 years ago
5 0

Given:

The two functions are:

f(x)=x^2

g(x)=\left(\dfrac{1}{3}x\right)^2

To find:

The statement that best compares the graph of g(x) with the graph of f(x).

Solution:

The horizontal stretch is defined as:

g(x)=f(kx)            ...(i)

If 0, the function f(x) is horizontally stretched by factor \dfrac{1}{k}.

If k>1, the function f(x) is horizontally compressed by factor \dfrac{1}{k}.

We have,

f(x)=x^2

g(x)=\left(\dfrac{1}{3}x\right)^2

Using these functions, we get

g(x)=f\left(\dfrac{1}{3}x\right)         ...(ii)

On comparing (i) and (ii), we get

k=\dfrac{1}{3}

Since 0, the function f(x) is horizontally stretched by factor \dfrac{1}{\frac{1}{3}}=3.

Hence, the correct option is D.

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Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u>p_1<u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u>p_2<u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no significant difference in seat belt use}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

                     T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers 16-24 years old who buckle up = 79%

\hat p_2 = sample proportion of drivers 25-69 years old who buckle up = \frac{924}{1100} = 84%

n_1 = sample of 16-24 years old drivers = 1000

n_2 = sample of 25-69 years old drivers = 1100

So, <u><em>test statistics</em></u>  =  \frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }  

                              =  -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that there is a difference in seat belt use.

4 0
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