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Setler [38]
3 years ago
9

Which rule has a scale factor of 3 and shifts the shape 4 units left and 2 units down?

Mathematics
1 answer:
Yanka [14]3 years ago
8 0

Answer: option b

Step-by-step explanation:

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Kyle and Michael run a lap around a track. Kyle gets a 50 meter head start and runs 3 meters per second. Michael runs at a speed
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A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores
EastWind [94]

Answer:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

Step-by-step explanation:

We have the following info given from the problem

\bar X_1 = 52 sample mean for this year

s_1= 13 sample deviation for this year

n_1 = 75 random sample selected for this year

\bar X_2 = 49 sample mean for last year

s_2= 11 sample deviation for last year

n_1 = 53 random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

(\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 = 75+53-2= 126

The confidence level is 0.95 and the significance would be \alpha=0.05 and \alpha/2 =0.025 so then the critical value for this case is :

t_{\alpha/2}= 1.979

The margin of error would be:

ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300

And the confidence interval would be given by:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

6 0
3 years ago
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