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Anit [1.1K]
3 years ago
14

Can someone please help me with math.

Mathematics
1 answer:
iragen [17]3 years ago
5 0

Answer:

Answer C 2,799.159

Step-by-step explanation:

v=\pi r^{2} h

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Which ratio correctly compares 120 cm to 80 mm?
bixtya [17]
120 cm=120 cm*(10 mm/1cm)=1200 mm,

Ratio:   1200 mm / 80 mm=15/1

Answer:

C.     15 to 1
3 0
3 years ago
Please help lol ............
Lyrx [107]

Answer:

1 X 16 and 8 X2

Step-by-step explanation:

1 X 16 = 16

2 X 8 = 16

3 X nothing = 16

4 X 4 = 16

5 X nothing = 16

6 X nothing = 16

7 X nothing = 16

8 X 2 = 16

etc.

8 0
3 years ago
Read 2 more answers
What is the slope of the line that passes through the points (-2,12) and (0,3)
ycow [4]

Answer:

-9/2

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(3-12)/(0-(-2))

m=-9/(0+2)

m=-9/2

4 0
2 years ago
A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after
lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                         P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of persons who were still employed primarily as engineers  = \frac{111}{200} = 0.555

           n = sample of graduates = 200

           p = population proportion of engineers

<em>Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                 significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } , 0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } ]

 = [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0
3 years ago
Find the distance between the points (7, -4) and (7,5).
Nadusha1986 [10]

Answer:

9

Step-by-step explanation:

find absolute value

5 0
2 years ago
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