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3 years ago

How do you do question 22

Mathematics

1 answer:

Alborosie3 years ago

6 0

Answer:

∠YXZ = 58°

Step-by-step explanation:

Use the angle sum theorem and the given relations.

∠WXY +∠WXV = ∠YXV

∠WXY +(11° +3∠WXY) = 139°

4∠WXY = 128° . . . . . subtract 11°

∠WXY = 32° . . . . . . . divide by 4

Similarly, ...

∠WXY +∠YXZ = 90°

∠YXZ = 90° -∠WXY = 90° -32°

∠YXZ = 58°

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PLZ IM BEING TIME What is the reciprocal of 6 1/4 ?

Soloha48 [4]

Answer:

4/25

Step-by-step explanation:

6 1/4=25/4

The reciprocal of 25/4 is 4/25.

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3 years ago

Which of the following are solutions to | x-1|= 5x + 2 ? Check all that apply.

agasfer [191]

By definition of absolute value, if $x-1\ge0$ , then $|x-1|=x-1$ , so

$|x-1|=5x+2\implies x-1=5x+2\implies4x=-3\implies x=-\dfrac34$

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3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$