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Marina86 [1]
3 years ago
7

Find the slope and y-intercept of the line 6x + 2y = –88

Mathematics
1 answer:
ira [324]3 years ago
4 0
Y=mx+b
6x+2y=-88
2y=-88-6x
y=-44-3x
y=-3x-44
m is the slope
b is the y-intercept
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An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ
Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558

b) The approximate probability that at least 9 carry the gene is:

P(x\geq9)=1-P(x\leq 8)\\\\\\

P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\

P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021

8 0
3 years ago
If E F = 7 , A C = 16 , and A F = 9
Vilka [71]

Answers:

CB = 14

GF = 8

FB = 9

EF is parallel to CB

====================================

Explanations:

Points E and F are midpoints of their respective sides. They form the midsegment EF. Because EF is a midsegment, A midsegment is half the length of its parallel counterpart, so CB is two times longer than EF. If EF is 7 units long, then CB = 2*EF = 2*7 = 14

For similar reasons, GF is parallel to AC. If AC = 16, then half of that is GF = (1/2)*AC = 0.5*16 = 8.

FB = FA = 9 as these segments have the same single tickmark to indicate they are the same length

EF is parallel to CB because EF is a midsegment, and this is one of the properties of being a midsegment. We can show that quadrilateral EGBF is a parallelogram to help prove this.

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Answer:

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Step-by-step explanation:

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Answer:

You have to have a Y and X axis line it may help. You can also time the few and yeah your smart you can do this i can help with more.

Step-by-step explanation:

I do this work. Thanks have a splendid life! Ask 4 me to solve more! Byes!

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