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3 years ago

Standard for ola (y + 3)2 = -12(x - 2), iden 1. D n111

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

Answer:

See below

Step-by-step explanation:

Your question is a bit unclear, so I'm going to assume you want the value of y or x:

(y+3)^2=-12(x-2)

-(y+3)^2/12=x-2

-(y+3)^2/12+2=x

(y+3)^2=-12(x-2)

y+3=sqrt(-12x+24)

y=sqrt(-12x+24)-3

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Donna is planning to take a trip to Spain that will cost $2,927. She has already saved $1,852. To earn the rest of the money, Do

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Answer:

2x + y = 17

-6x = 3y - 51

Step-by-step explanation:

You get an infinite number of solutions when two equations are really just the same. This means you can convert one into the other simply by multiplying one to get the other.

In the above answer, if you multiply the above by -3 you get:

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3 years ago

“Jessie has 2/3 of a bag of apples that she wants to divide equally among her 5 siblings. What fractional part of the bag

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3 years ago

Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask

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Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$

$=0.6^8+8(0.4)(0.6)^7+28(0.4)^2(0.6)^6+56(0.4)^3(0.6)^5+70(0.4)^4(0.6)^4+56(0.4)^5(0.6)^3$

$=0.95019264$

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3 years ago