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Bad White [126]
3 years ago
10

Question 3 5 1 Divide 6 • 12

Mathematics
2 answers:
rosijanka [135]3 years ago
8 0

Answer:

10

Step-by-step explanation:

(5/6)÷(1/12) = 10

puteri [66]3 years ago
3 0

Answer:

5/6 divided by 1/12 = 10.

Hope this helped!

Step-by-step explanation:

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We draw a card from a standard deck. What is the probability that it is a Jack, Queen, King or it is red.
enyata [817]

Answer:

8/13 chance to get a jack, Queen, King or Red

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an airline experiences a no show rate of 6%. what is the maximum number of reservations that it could accept for a flight with a
Alborosie
168 passengers because 168*95%=159.6

Therefore 168 passengers would have to book the flight
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3 years ago
4 Compare 5.269 and 5.038.
Hunter-Best [27]

Answer:

the answer is B

Step-by-step explanation:

4 0
3 years ago
how many pounds of coffee worth $10 a pound should be added to 20 pounds of coffee worth $4 a pound to get a mixture worth $5 a
Solnce55 [7]

Answer:

4

Step-by-step explanation:

1. if the mass of coffee worth $10 is 'x', then its worth is 10x, and the worth of the mixture is 5(x+20);

2. using the previous item it is possible to write the next equation:

5(x+20)=4*20+10*x; ⇒ x=4.

3. the required mass is 4 pounds.

5 0
3 years ago
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
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