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ycow [4]
3 years ago
6

Can someone help me with this

Mathematics
1 answer:
svp [43]3 years ago
5 0

I have added the solution, please specify as a comment if you do not understand..

r = 9 ft

h = 12 ft

Answer is 1186,92

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dedylja [7]
1:5 is the answer in 1:n form
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3 years ago
F(-12) if f(x) = 2x + 15
Alenkinab [10]

Answer: f(-12)=-9

Step-by-step explanation:

We have the following function:

f(x)=2x+15

And we have to evaluate it when x=-12, this means we have to substitute x by -12:

f(-12)=2(-12)+15

Solving:

f(-12)=-24+15

Finally:

f(-12)=-9

8 0
4 years ago
Can someone teach me how to balance algebra equations, I just don't get how to do it. Like with fractions?
natka813 [3]

Answer:

<h3>Move all the x terms to one side. Use inverse operations and add 1 5 x 15x 15x to both sides to keep the equation balanced. Solve by working backwards from the order of operations. This means we need to undo the −2 first by adding 2 to both sides of the equation to keep it balanced.</h3>

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hichkok12 [17]

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6 0
3 years ago
A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same di
Lena [83]

Answer:

Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .

Step-by-step explanation:

We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;

Mean, \mu = 1.3 cm            and           Standard deviation, \sigma = 0.01 cm

Also, since distribution is normal;

                 Z = \frac{X -\mu}{\sigma} ~ N(0,1)

Let X = range of diameters

So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)

  P(X < 1.33) = P( \frac{X -\mu}{\sigma} < \frac{1.33 -1.3}{0.01} ) = P(Z < 3) = 0.99865

  P(X <= 1.27) = P( \frac{X -\mu}{\sigma} < \frac{1.27 -1.3}{0.01} ) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865

                                                                                            = 0.00135

 P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .

Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .

7 0
3 years ago
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