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3 years ago

9

Find angle N and arc NQ. See the image below.

1 answer:

malfutka [58]3 years ago

3 0

Answer:

Angle N=32

Arc NQ=106

Step-by-step explanation:

Angle N is a inscribed angle of Arc MP so that means Angle N measures 32.

Angle NPQ is a inscribed angle of Arc NQ so that means Arc NQ is twice the measures of NPQ so

Arc NQ=106

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Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 8 sin(x

V125BC [204]

Answer with explanation:

The equation of two curves are

y=8 sin x

y=8 cos x

The point of intersection of two curves are

→8 sin x=8 cos x

sinx = cos x

$x=\frac{\pi}{4}\\\\y=8\sin\frac{\pi}{4}\\\\y=4\sqrt{2}$

When you will look between points , 0 and $x=\frac{\pi}{4}\\\\y=8\sin\frac{\pi}{4}\\\\y=4\sqrt{2}$ ,the curve obtained is right angled triangle.

Now, rotate the curve that is right angled triangle along the line , y= -1, to obtain a shape similar or resembling with Right triangular prism.

Draw a circular disc in the right triangular prism , having radius =x,and small part on the triangle having length dx.dx varies from 0 to $\frac{\pi}{4}$ .

Required volume of solid obtained

$=\int\limits^{\frac{\pi}{4}}_0 {8 \sin x} \, dx \\\\=|-8\cos x|\left \{ {{x=\frac{\pi}{4}}} \atop {x=0}} \right.\\\\=8-4\sqrt{2}$

=8-4√2 cubic units

5 0

3 years ago

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

4 0

3 years ago

Simplify (-8q^3r^4s^2)^2

koban [17]

Answer:

64q^6r^8s^4

Step-by-step explanation:

(-8q^3r^4s^2)^2

= (-8)^2 q^(3*2) r^(4*2) s^(2*2)

= 64q^6r^8s^4

3 0

3 years ago

Which set of ordered pairs represents y as a function of X? a{(-9, 2), (0, 6), (1, -2), (-3, 6)} b{(-1,0), (4, 3), (-7, -3), (-1

olga nikolaevna [1]

A is your answer

Explanation:

When dealing with functions, x’s never repeat.

8 0

3 years ago

Find de product of 15 and 107

Andru [333]

Answer:

$15 \times 107 = 1605$

6 0

3 years ago

Read 2 more answers