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brilliants [131]
2 years ago
15

Given the parent function f(x)=x^2 give the best description of the graph of y=x^3+3. a. expanded vertically by a factor of 5 c.

expanded horizontally by a factor of 5 b. expanded vertically by a factor of 10 d. expanded horizontally by a factor of 10
Mathematics
2 answers:
4vir4ik [10]2 years ago
8 0

Answer:

Step-by-step explanation:nsnsnmalqlq

padilas [110]2 years ago
4 0

Answer: We have the following function:

f (x) = x ^ 2

We have the following transformation:

Expansions and horizontal compressions

The graph of y = f (bx):

If 0 <b <1, the graph of y = f (x) expands horizontally by the factor of 1 / b.

Applying the transformation:

y = (0.2x) ^ 2

The factor is:

1 / b = 1 / 0.2 = 5

Answer:

b. expanded horizontally by a factor of 5

Step-by-step explanation:

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Answer:$2.88

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Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist
satela [25.4K]

Answer:

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

p_v =2*P(t_{(9)}>1.539)=0.158  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

Step-by-step explanation:

Data given and notation  

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

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

And the sample deviation given by:

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

And we got:

\bar X=7.55 represent the sample mean

s=0.103 represent the sample standard deviation

n=10 sample size  

\mu_o =7.5 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 (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 7.5  

Alternative hypothesis:\mu \neq 7.5  

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{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

P-value

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

df=n-1=10-1=9  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(9)}>1.539)=0.158  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

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Step-by-step explanation:

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Step-by-step explanation:

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<h3>stay safe healthy and happy.</h3>
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