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Delicious77 [7]
3 years ago
12

A company that manufactures memory chips for digital cameras uses the formula c equals 120 cube root of n squared end root plus

1300to determine the cost c, in dollars, of producing n chips.
How much will it cost to produce 250 chips?
Mathematics
2 answers:
Galina-37 [17]3 years ago
8 0

Answer:

$6062

Step-by-step explanation:

The formula we are given is

c=120\sqrt[3]{n^2}+1300

We will replace n with 250:

c=120\sqrt[3]{250^2}+1300\\ \\=120\sqrt[3]{62500} +1300\\\\=120(39.685)+1300\\\\=4762.20+1300\\\\=6062.20 \approx 6062

steposvetlana [31]3 years ago
6 0
The cost:C = 120 · ∛(n²) + 1,300We have to produce 250 memory chips for digital cameras:n = 250C ( 250 ) = 120 · ∛(250²) + 1,300 = = 120 · ∛62,500 + 1,300 = = 120 · 39.685 + 1,300 = = 4,762 + 1,300 = $6,062Answer: It will cost $6,062.

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the coefficient is A.5

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Where does the section of the bridge meet ground level? Solve 0 = -x2 + 10x - 8
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Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
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

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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
3 years ago
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hodyreva [135]

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7 0
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