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kari74 [83]
3 years ago
14

Solve for x: −2(x − 2)2 + 5 = 0

Mathematics
2 answers:
Dimas [21]3 years ago
6 0

Answer: x = 3.58, 0.42

Step-by-step explanation: i took the test lol

Licemer1 [7]3 years ago
5 0

Option A

The values of x are 3.58 or 0.42

<u>Solution:</u>

Given, equation is -2(x-2)^{2}+5=0

We have to solve for the x and choose from the options given,

Now, take the equation,

\begin{array}{l}{\rightarrow-2(x-2)^{2}+5=0} \\\\ {\rightarrow-2(x-2)^{2}=-5}\end{array}

Negative sign cancels on both sides

\begin{array}{l}{\rightarrow 2(x-2)^{2}=5} \\\\ {\rightarrow(x-2)^{2}=\frac{5}{2}}\end{array}

Taking square root on both sides, we get

\begin{array}{l}{\rightarrow x-2=\sqrt{\frac{5}{2}}} \\\\ {\rightarrow x-2=\pm 1.58} \\\\ {\rightarrow x=2 \pm 1.58}\end{array}

\begin{array}{l}{\rightarrow x=2+1.58 \text { or } 2-1.58} \\\\ {\rightarrow x=3.58 \text { or } 0.42}\end{array}

Hence, the values of x are 3.58 or 0.42, so option A is correct.

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A company maintains a large fleet of automobiles. To check the average number of miles driven per month, a random sample of 100
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Answer:

df=n-1=100-1=99  

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

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Step-by-step explanation:

Data given

\bar X=2690 represent the sample mean

s=360 represent the sample standard deviation

n=100 sample size  

\mu_o =2600 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)  

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We need to conduct a hypothesis in order to check if the true mean of interest is different from 2600, the system of hypothesis would be:  

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Alternative hypothesis:\mu \neq 2600  

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Calculate the statistic

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P-value

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

df=n-1=100-1=99  

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

p_v =2*P(t_{(99)}>2.7)=0.0082  

And since the p vaue is lower than the significance level we have enough evidence to reject the null hypothesis.

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