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Eddi Din [679]
3 years ago
15

Can someone pls solve this for me

Mathematics
1 answer:
Ludmilka [50]3 years ago
6 0

Given:

m∠WYX = 20°

m∠XYZ = 135°

To find:

The measure of arc WXZ.

Solution:

m∠WYZ = m∠WYX + m∠XYZ

m∠WYZ = 20° + 135°

m∠WYZ = 155°

The measure of inscribed angle is half of the intercepted arc.

$\Rightarrow m \angle WYZ = \frac{1}{2} m (ar WXZ)

$\Rightarrow 155^\circ = \frac{1}{2} m (ar WXZ)

Multiply by 2 on both sides.

$\Rightarrow 2\times 155^\circ = 2 \times  \frac{1}{2} m (ar WXZ)

$\Rightarrow 310^\circ = m (ar WXZ)

The measure of arc WXZ IS 310°.

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Answer:

\frac{ \sqrt{6} }{2}

Step-by-step explanation:

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Basically we want to know how many times 5/8 goes into 5 1/4.

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Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
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Answer:

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