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IgorLugansk [536]
2 years ago
8

I need to simplfiy the answer can someone help me with the question

Mathematics
1 answer:
erica [24]2 years ago
7 0

Answer:

q

Step-by-step explanation:

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manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasin
alexdok [17]

Answer:

P(X>17) = 0.979

Step-by-step explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r) nCr q^{r} p^{n-r}

nCr = \frac{n!}{(n-r)!r!}

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) = 20C18 * 0.97^{18} * 0.03^{20-18}

P(X=18) = 20C18 * 0.97^{18} * 0.03^{2}

P(X=18) = 0.0988

P(X=19) = 20C19 * 0.97^{19} * 0.03^{20-19}

P(X=19) = 20C19 * 0.97^{19} * 0.03^{1}

P(X=19) = 0.3364

P(X=20) = 20C20 * 0.97^{20} * 0.03^{20-20}

P(X=20) = 20C20 * 0.97^{20} * 0.03^{0}

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch

6 0
3 years ago
Factorise completely:<br> 12a+8b
zavuch27 [327]
12a + 8b....common factor is 4
4(3a + 2b) <==
6 0
3 years ago
<img src="https://tex.z-dn.net/?f=12%20%5Cfrac%7B3%7D%7B6%7D%20%2B%2014%5Cfrac%7B4%7D%7B6%7D%20" id="TexFormula1" title="12 \fra
Vikki [24]
27 1/6 or 163/6 or 27.16
6 0
3 years ago
Read 2 more answers
Anderson spends more than $25 to buy 5 cans of vegetable broth and some pounds of beans. The cost of 5 cans of vegetable broth i
Lera25 [3.4K]
If you looking for the amount of beans

25 = 16.75 + 1.50×b

25-16.75 = 1.50b

8.25 = 1.50b

8.25/1.50 = b

b = 5.5
7 0
3 years ago
Calculate the average rate of change of f(x) = 1 x - x2 - 2 for 3 ≤ x ≤ 6. A) -163 18 B) -18 163 C) 163 18 D) 18 163
larisa86 [58]
To solve this we are going to use the average rate of change formula: A(x)= \frac{f(b)-f(a)}{b-a}
where
A(x) is the average rate of change of the function
f(a) is the position function evaluated at a
f(b) is the position function evaluated at b
a is the first point in the interval
b is the second point in the interval

We can infer for our problem that the first point is 3 and the second point is 6, so a=3 and b=6. Lets replace those values in our formula:
A(x)= \frac{f(b)-f(a)}{b-a}
A(x)= \frac{f(6)-f(3)}{6-3}
A(x)= \frac{6-6^2-2-(3-3^2-2)}{3}
A(x)= \frac{-32-(-8)}{3}
A(x)= \frac{-32+8}{3}
A(x)= \frac{-24}{3}
A(x)=-8

We can conclude that the average rate of change of the function f(x) = 1 x - x2 - 2 for <span>3 ≤ x ≤ 6 is -8</span>
3 0
3 years ago
Read 2 more answers
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