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PolarNik [594]
2 years ago
11

Which phrase best describes the translation from the graph y = 2(x – 15)2 + 3 to the graph of y = 2(x – 11)2 + 3? 4 units to the

left 4 units to the right 8 units to the left 8 units to the right
Mathematics
2 answers:
Mama L [17]2 years ago
8 0
Left is plus right is negative -11-(-15)=4  so you know 4 to the left
ira [324]2 years ago
4 0

we have

y=2(x-15)^{2}+3

this is the equation of a vertical parabola open up with vertex at point (15,3)

y=2(x-11)^{2}+3

this is the equation of a vertical parabola open up with vertex at point (11,3)

so

the rule of the translation is

(x,y)---------> (x-4,y)

that means

the translation is 4 units to the left

therefore

<u>the answer is</u>

4 units to the left

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The average life of a certain type and brand of battery is 75 weeks. The average life of each of 9 randomly selected batteries i
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Answer:

Option A

Step-by-step explanation:

Since at. 0.05 level of significance, the p value was great than 0.05 but less than 0.1, we will fail to reject the null and conclude that the sample data does not provide enough evidence that the average life of the batteries is greater than 75 weeks.

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Answer:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

\bar X_1= 75.1 represent the sample mean for the scores of the undergraduate students

s_1 = 12.8 represent the standard deviation for the undergraduate students

n_1 =35 the sample size for the undergraduate

\bar X_2= 72.1 represent the sample mean for the scores of the high school students

s_2 = 14.6 represent the standard deviation for the high school students

n_2 =50 the sample size for the high school

The confidence interval for the true difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df=n_1 +n_2 -2= 35+50-2=83

The confidence level is 90% and the significance level is \alpha=0.1 and \alpha/2 =0.05 then the critical value would be:

t_{\alpha/2}= 1.99

And replacing the info we got:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

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