Can someone help me? its simplify.. <br>

Help Me Pls Thank You

mart [117]

Answer: It's b

Step-by-step explanation:

7 0

3 years ago

What irrational number is between 4.12 and 4.13<br><br> A.)sqrt{19}<br> B.)sqrt{23}

fenix001 [56]

Sort{17} which equals 4.123

3 0

3 years ago

Factor the trinomial x^2 - 2x - 8

ValentinkaMS [17]

Well what two numbers multiply to get -8 and add to get -2?
Well, to get -8, we either have (-1,8),(-2,4),(-4,2),(-8,1)
8-1=7
-2+4=2
2-4=-2
1-8=-7
Thus it should be (-4,2),
This means we should get (X-4)(X+2)

7 0

3 years ago

I forgot how to do most of this yet they gave it to me as hw ‍♀️

svp [43]

18.

-2(-9)+3

18+3

21

Explanation: -2(-9) When multiplying numbers that have both a negative sign, the answer is positive. Multiply normally.

19.

-6(-4)0

24(0)

0

Explanation: -6(-4) As I explained before, the answer would be positive.

24(0) Every number multiplied by 0 is 0.

20.

56+(-7)

49

Explanation: 56+(-7) Any problem like this is just basically going to be 56-7.

21.

-36+(-4)

-40

Explanation: -36+(-4) You are simply adding normally then adding the sign. This will only work if the signs are the same. (Signs are positive and negative.)

8 0

3 years ago

Read 2 more answers

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

Can someone help me? its simplify.. ​

Can someone help me? its simplify..