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stellarik [79]
2 years ago
10

Question (e) only please show working as well appreciate it

Mathematics
1 answer:
Georgia [21]2 years ago
3 0

Answer:

a. f(2) = 6

b. f(3) + f(-3) = 62

d. 3(x+1)^2 - 5(x+1) + 4

e. This one looks like the secant line equation. Not fully sure.

Step-by-step explanation:

You replace whatever is inside of the parenthesis with the x in the function. For the first two, you just have to calcluate the numbers.

For the other two , you have use algebra. For d, you replace x with x+1.

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Answer:

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2 years ago
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

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The p value for this case would be given by:

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Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0
3 years ago
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poizon [28]
Factor 5 de 5t - 5
5(t - 1)

¡Espero que esto ayude! :)
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Answer:

about 700

Step-by-step explanation:

you add 204 + 498=

4 0
3 years ago
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