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Lera25 [3.4K]
3 years ago
11

Jill's bowling scores are approximately normally distributed with mean 170 and standard deviation 25, while jack's scores are ap

proximately normally distributed with mean 160 and standard deviation 10. if jack and jill each bowl one game, then assuming their scores are independent random variables, find the probability that
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0
Jill's bowling scores<span> are </span>approximately normally distributed<span> with </span>mean 170<span> and </span>standard deviation<span> 20,</span>while Jack's scores<span> are ... </span>If Jack<span> and </span>Jill each bowl one game<span>, </span>then assuming<span> that </span>their scores<span> are ... The difference of </span>means<span> of </span>Jack<span> and </span>Jill<span> follows a </span>normal distribution<span> with </span>mean<span> -</span>10<span> and </span>standard deviation<span> is: </span>25<span>.</span>
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7.
Iteru [2.4K]

Answer:

a. 45 π

b. 12 π

c. 16 π

Step-by-step explanation:

a.

If a 3×5 rectangle is revolved about one of its sides of length 5 to create a solid of revolution, we can see a cilinder with:

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Height: 5

Then the volume of the cylinder is:

V=π*r^{2} *h= π*(3)^{2} *(5) = π*(9)*(5)=45 π

b. If a 3-4-5 right triangle is revolved about a leg of length 4 to create a solid of revolution. We can see a cone with:

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Then the volume of the cone is:

V=(1/3)*π*r^{2} *h= (1/3)*π*(3)^{2} *(4) = (1/3)*π*(9)*(4)=12 π

c. We can answer this item using the past (b. item) and solving for the other leg revolution (3):

Then we will have:

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3 years ago
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2 years ago
(a) By inspection, find a particular solution of y'' + 2y = 14. yp(x) = (b) By inspection, find a particular solution of y'' + 2
SOVA2 [1]

Answer:

(a) The particular solution, y_p is 7

(b) y_p is -4x

(c) y_p is -4x + 7

(d) y_p is 8x + (7/2)

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 14.

Because the nonhomogeneus part of the differential equation, 14 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 14

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 14

A = 6/2 = 7

Therefore, the particular solution, y_p = A is 7

(b) y'' + 2y = −8x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x

2Ax + 2B = -8x

By inspection,

2B = 0 => B = 0

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x

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Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x + 14

2Ax + 2B = -8x + 14

By inspection,

2B = 14 => B = 14/2 = 7

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x + 7

(d) Find a particular solution of y'' + 2y = 16x + 7

Let y_p = Ax + B

y'_p = A

y''_p = 0

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2Ax + 2B = 16x + 7

By inspection,

2B = 7 => B = 7/2

2A = 16 => A = 16/2 = 8

The particular solution y_p = Ax + B

is 8x + (7/2)

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3 years ago
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