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Vera_Pavlovna [14]
3 years ago
12

A training company provides classroom and online instruction to hundreds of people per month. As part of their training program

a satisfaction survey is conducted at the end of each course. The training company believes there might be a change in its satisfaction rates. During the prior calendar year their satisfaction rate was 94.2%. Over the past 6 months the number of satisfied responses was 824 out of 892 surveys. Use the information provided to perform a 1 sample proportion test to determine if there is a difference between the current satisfaction rate and the prior year.

Mathematics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

According to the sample data, we can claim that the current satisfaction rate is different from last year's.

Step-by-step explanation:

The explanation is in the picture.

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Answer:

18/5

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3 years ago
Which is the slope of the line y=−3x+2? −3 −2 2 3
Gnesinka [82]

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-3

Step-by-step explanation:

y=mx+b where m=slope and b=y-intercept.

3 0
3 years ago
Hhelp pls lol i don't understand​
sladkih [1.3K]

Answer:

m∠1= 80

m∠2= 100

m∠3= 80

Step-by-step explanation:

Given: m∠1= 2x +40

m∠2= 2y +40

m∠3= x +2y

m∠1 +m∠2= 180 (adj. ∠s on a str. line)

2x +40 +2y +40= 180

2x +2y+ 80= 180

2x +2y= 180 -80

2x +2y= 100

2(x +y)= 100

x +y= 100 ÷2

x +y= 50 -----(1)

m∠1= m∠3 (vert. opp. ∠s)

2x +40= x +2y

2x -x +40= 2y

x= 2y -40 -----(2)

Substitute equation (2) into (1):

2y -40 +y= 50

3y= 50 +40

3y= 90

y= 90 ÷3

y= 30

Substitute y= 30 into equation (2):

x= 2(30) -40

x= 60 -40

x= 20

m∠1

= 2x +40

= 2(20) +40

= 40 +40

= 80

m∠2

= 2y +40

= 2(30) +40

= 60 +40

= 100

m∠3

= x +2y

= 20 +2(30)

= 20 +60

= 80

Alternatively, since m∠1= m∠3,

m∠3

= m∠1 (vert. opp. ∠s)

= 80

5 0
3 years ago
How many student preferred math??
stich3 [128]
B,13 students prefered math
5 0
3 years ago
Perimeter/area (triangles)
Rzqust [24]

Answer:

12 ft²

50 ft²

Step-by-step explanation:

∆ABD = ∆CBD

Perimeter of ∆ABD = 12 ft²

Perimeter of triangle ∆CBD

If ∆ABD = ∆CBD

Then, Perimeter of ∆CBD = 12 ft²

∆ABC = ∆CDA

Area of ∆ABC = 25 ft²

Then the area of rectangle ABCD = Area of ∆ABC + Area of ∆CDA

= 25 ft² + 25 ft²

= 50 ft²

6 0
2 years ago
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