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zaharov [31]
3 years ago
12

Consider a solid spherical ball made of wood. Suppose a hole is bored (drilled) vertically through the center of the ball and th

e resulting solid has a height of 8 inches. What is the volume of the resulting solid?
Mathematics
1 answer:
lbvjy [14]3 years ago
5 0

Answer:

The volume of solid is 268.08 inches cube.

Step-by-step explanation:

The solid spherical ball has height 8 inches that means the diameter of the sphere is 8 inches. So, the radius will be the half of diameter. Thus, radius is 4 inches.

Now use below formula to find the volume of solid sphere.

Volume = \frac{4}{3}\pi r^{2} \\Volume = \frac{4}{3}\pi (4)^{2} \\Volume = 268.08 \ in^{3}

Therefore, the volume of solid is 268.08 inches cube.

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22 movies per month.

Step-by-step explanation:

30=9.99+0.89m

30-9.90=0.89m

20.01/0.89=m

22.48=m

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At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective pa
11Alexandr11 [23.1K]

Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 7 parts

            r = number of success = fewer than 2

           p = probability of success which in our question is % of defective

                 parts produced by one of the machine, i.e; 19%

<em>LET X = Number of parts that are defective</em>

<u>So, it means X ~ Binom(n = 7, p = 0.19)</u>

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

    P(X < 2) = P(X = 0) + P(X = 1)

                  =  \binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}

                  =  1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}

                  =  <u>0.604</u>

<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>

8 0
2 years ago
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