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zimovet [89]
3 years ago
15

Daniel hosted a party in which 15 to 45 people could attend and he charged $7 per person for admission. What is the range of adm

ission with respect to the number of people that attended the party?
A. (105,315)
B. [15,45]
C. [105,315]
D. (15,45)​
Mathematics
1 answer:
irina [24]3 years ago
6 0

Answer:

c. [105,315]

Step-by-step explanation:

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Please help! Thank you so much.
inysia [295]

Answer:

4 over 5

Step-by-step explanation:

its 4 over 5 because your going to get a decimal and the other answers you will get a integer which the question isnt asking for so the answer is 4 over 5

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3 years ago
HELPPPPP!!! Can you TEACH ME HOW TO DO THIS and will give b
IRINA_888 [86]

Answer:

yh I can help the answer is 185/1000 bc u have to convert it as a fraction and u can simplify it by 5 then the answer will be 37/200 plz try this if not I will tell u something else

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3 years ago
Frances read 4/5 of an article in three minutes. How much of the article did she read each minute?
Margarita [4]

Answer:

0.267

Step-by-step explanation:

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8 0
2 years ago
HELP ASAP DUE TOMORROW!!!!
Vikentia [17]

Answer:

c

Step-by-step explanation:

3 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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