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likoan [24]
3 years ago
10

Please answer this correctly

Mathematics
1 answer:
Dahasolnce [82]3 years ago
8 0

Answer:

9 5/7

Step-by-step explanation:

You add all the 13/14 then divided it by 14

Then add all the while numbers then add then add them all together

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A truck with 36-in.-diameter wheels is traveling at 55 mi/h.
kkurt [141]

Answer:

The angular speed is 3,227 rad/min

Step-by-step explanation:

Remember that

1 mile=63,360 inches

step 1

Find the circumference of the wheels

The circumference is equal to

C=\pi D

we have

D=36\ in

substitute

C=\pi (36)

C=36\pi\ in

step 2

we know that

The speed of the wheel is 55 mi/h

<em><u>Convert to mi/min</u></em>

55 mi/h=55/60 mi/min

<u><em>Convert to in/min</em></u>

(55/60) mi/min=55*63,360/60 in/min= 58,080 in/min

we know that

The circumference of the wheel subtends a central angle of 2π radians

so

using proportion

Find out how much radians are 58,080 inches

\frac{36\pi }{2\pi }=\frac{58,080}{x} \\\\x=2*58,080/36\\\\x=3,226.67 \ rad

therefore

The angular speed is 3,227 rad/min

6 0
3 years ago
Which accounts of a business can outsiders view
Ainat [17]
Non gap businesses are allowed to be viewed by outsiders<span />
6 0
4 years ago
What is the area of this triangle?<br><br> Enter your answer in the box.
olga2289 [7]

Answer: I'm pretty sure its 12

Step-by-step explanation: You count the unit length of the base and then find the height. Now you multiply them and divide by 2.

Hope this helps.

Branliest please

7 0
3 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
4 years ago
How is a 15% discount similar to a 15% decrease? Explain.
andrey2020 [161]
They are the same. Example: If something costs 100$ and you found a coupon to get a 15% discount you will save: 15$ and your cost will be 100$-15$=85$ Hope this helped! Have a great day!

7 0
3 years ago
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