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Ksenya-84 [330]
3 years ago
10

Y-1=4y-2/3 y=_____ solve

Mathematics
2 answers:
Vikki [24]3 years ago
7 0
Y - 1  =  4y - 2/3.      Move the ys over to one side,  +4y crosses over to become -4y,
                                  and -1 crosses over the right side to become +1.
y -4y  = -2/3  + 1
-3y  =    1 - 2/3
-3y =  1/3          Divide both sides by -3.

-3y/-3  = (1/3) / -3.
y  =  1/3  * -1/3
y =  - 1/9

Alexxandr [17]3 years ago
4 0


y-1=4y-2/3

<em><u>-4y        -4y        </u></em>

<em>-</em>3y-1=-2/3

<em><u>+1          +1</u></em>

-3y=1-2/3

-3y=3/3-2/3

-3y=1/3      <em>/:(-3)</em>

y=1/3*(-1/3)

y=-1/9

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Answer: <em>6.6 meters a week</em>

Step-by-step explanation:

<em>This problem is quite simple</em>

<em>You take 80 and divide it by 12</em>

<em>80/12 </em>

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25x^2 -1

Step-by-step explanation:

If you multiply parentheses:

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25x^2 - 5x + 5x -1

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A simpler way is to memorize this formula

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Read 2 more answers
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
3 years ago
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