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3 years ago

14

luis earns a base salary of $150.00 every week with an additional 14% commision on everything he sells. luis sol $6050.00 worth

of items last week. What was luis's total pay last week

2 answers:

kodGreya [7K]3 years ago

7 0

Answer:

$6897

Step-by-step explanation:

14% of $6050 is $847

Then add 847 to 6050 and you get 6897

vfiekz [6]3 years ago

6 0

Answer:

$992.8

Step-by-step explanation:

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Plz help me out in that because i don’t what is the value of 5n+9+n

valentina_108 [34]

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3 years ago

Kay [80]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.535 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.2 \leq \sigma \leq 2.8$

And the best option would be:

A. 2.2 < σ < 2.8

Step-by-step explanation:

Information provided

$\bar X=32.1$ represent the sample mean

$\mu$ population mean

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=83-1=82$

The Confidence is given by 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=62.132$

$\chi^2_{1- \alpha/2}=104.139$

And replacing into the formula for the interval we got:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.535 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.2 \leq \sigma \leq 2.8$

And the best option would be:

A. 2.2 < σ < 2.8

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antiseptic1488 [7]

90-48 = 42
3 units = 42
1 unit = 14
48-14=34

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