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MAVERICK [17]
3 years ago
13

EASY SUBTRACTING AND MULTIPLYING QUESTION (not 128)

Mathematics
1 answer:
Serga [27]3 years ago
3 0

Answer:

x(n)=64

Step-by-step explanation:

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marishachu [46]

Answer: IT EQUALS 2

Step-by-step explanation:

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3 years ago
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Jacky spent 53% of all her money to buy a computer game. How much did the game cost, if Jacky had $120 before buying the game?
Keith_Richards [23]
The game cost 63.6$, because 53% of 120 is 63.6
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Gavin needs $80 to buy a fish tank. He has saved $8 and plans to work as a babysitter to earn $9 per hour. Which inequality show
yuradex [85]
Gavin needs $80 to buy a fish tank. He has saved $8 and plans to work as a babysitter to earn $9 per hour. Which inequality shows the minimum number of hours, n, that Gavin should work as a babysitter to earn enough to buy the fish tank? 8 + 9n ≥ 80, so n ≥ 8 8 + 9n ≤ 80, so n ≤ 8 9n ≥ 80 + 8, so n ≥ 9.8 9n ≤ 80 + 8, so n ≤ 9.8
3 0
2 years ago
How many times larger is the 5 in 528, 394 as in 352, 107?
seraphim [82]

Answer: B 10 Times large

Step-by-step explanation:The first 5 is 500,000 And the second 5 is 50,000

500,000÷ 50,000= 10

7 0
2 years ago
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

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
2 years ago
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