Tiffany ate ¼ of a pizza. If there were 16 slices of pizza to begin with, how many slices did Tiffany eat?

Mathematics

2 answers:

PIT_PIT [208]2 years ago

7 0

Answer:

she ate 4 slices

Step-by-step explanation:

Ksenya-84 [330]2 years ago

3 0

Answer:

1/4 × 16

therefore she ate 4 slices

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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

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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:

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

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