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2 years ago

7

A nurse works for a temporary nursing agency. The starting hourly wages for the six different work locations are $12.50, $11.75,

$9.84, $17.67, $13.88, and $12.98. As the payroll clerk for the temporary nursing agency, find the median starting hourly wage.

1 answer:

Tomtit [17]2 years ago

8 0

100 dollars and 32 cents

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Can someone please help me

belka [17]

Answer:

D

Step-by-step explanation:

First we need to figure out how many people are going on the field trip

if each class has 27 people going and there are 6 classes, 6*27=162

162 ppl are going. If 3 buses can hold 48 ppl then 3*48=144

3 buses are not enough, if we were to bring in one more bus 4*48=192

there would be enough place for everyone

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3 years ago

What is the equation of the line that is parallel to the line y=2x+3 and passes through the point (-2,1)

Lerok [7]

Answer:

y= 2x+5

Step-by-step explanation:

The equation of a line is usually written in the form of y=mx+c, where m is the gradient and c is the y-intercept.

Parallel lines have the same gradient. Hence m=2.

Subst. m=2 into the equation:

y=2x +c

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1= 2(-2) +c

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c= 1 +4

c=5

Thus, the equation of the line is y= 2x +5.

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2 years ago

Add 0.269 + 44.9 = <br><br> plz plz plz help

Pani-rosa [81]

Answer:

The answer is...

Step-by-step explanation:

45.169

3 0

3 years ago

Read 2 more answers

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

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3 years ago

Y=\sqrt(-x^(2)+2x) . tìm giá trị lớn nhất của hàm số

Jlenok [28]

Answer:

UM hello how are you im njot really gonna answer anywaysss byee

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2 years ago