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3 years ago

Compare the values of 2s and 5s in 55,220

1 answer:

4 0

Hello !
20= Twenty
200= two hundread
5,000= five thousand
50,000= fifty thousand
we always replace the numbers that comes after the value with 0's
Good luck

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Answer:

The answer is 63, but there is no option of the answer

Step-by-step explanation:

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2 years ago

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Determine the greatest common factor of the following numbers 8 and 12 NO LINKS PLEASE!!!!

jonny [76]

I think it would be 4.
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2 years ago

Which equation has a unit rate greater than the unit rate of the relationship shown in the graph?

shusha [124]

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3 years ago

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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:

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

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

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

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2 years ago

Figure ABCD - Figure EFGH

Colt1911 [192]

<h3>Answer:</h3>

30.5 units

<h3>Step-by-step explanation:</h3>

We know that:

Trapezoid ABCD has one side that is 25 unit.
The other 3 sides (AB, BC, and CD) are 12 units.
EF = 6 units
EH = 12.5
Trapezoid ABCD is congruent to Trapezoid EFGH

If the 3 sides of the trapezoid ABCD are the same sides, then side EF, FG, GH must be of the same length because of congruence. The value of FG and GH must be the same length as EF. We can clearly see in the picture that EF is 6 units. Hence, EF is 6 units, FG is 6 units, and GH is 6 units. The work of the perimeter is shown below.

Work

=> 6(3) + 12.5
=> 18 + 12.5
=> 30.5 units

Hence, the perimeter of EFGH is 30.5 units.

$BrainiacUser1357$

4 0

1 year ago