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stiv31 [10]
3 years ago
9

What are the partial products of 42×5

Mathematics
1 answer:
ale4655 [162]3 years ago
8 0
The partial would be
40x5=200
2x5=10
210
Your just breaking them apart
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3 years ago
During a promotion at Dale's Donuts, customers received
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divide 528 by 88 and you will get 6

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Step-by-step explanation:

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3 years ago
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3 years ago
In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol
uysha [10]

Answer:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

Step-by-step explanation:

For this problem we have the following data given:

\bar X_1 = 9 represent the sample mean for one of the departments

\bar X_2 = 8 represent the sample mean for the other department

n_1 = 25 represent the sample size for the first group

n_2 = 20 represent the sample size for the second group

s_1 = 2 represent the deviation for the first group

s_2 =1 represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true 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 confidence given is 95% or 9.5, then the significance level is \alpha=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n_1 +n_2 -2= 20+25-2= 43

And the critical value for this case is t_{\alpha/2}=2.02

And replacing we got:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

4 0
3 years ago
Round 57640 to the nearest ten thousands
stira [4]
60,000 because 7 is closer to 10.

4 0
3 years ago
Read 2 more answers
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