If Analissa puts 55 seeds into 11 pots, how many seeds does she put in 4 pots?

Mathematics

2 answers:

charle [14.2K]3 years ago

7 0

Answer: Since she puts 5 seeds per pot, I know that she will put 20 seeds in four pots.

I got this by dividing 55/11 to get the amount of seeds in one pot.

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Anton [14]3 years ago

6 0

She will put 20 in four pots

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3 years ago

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A national college researcher reported that 65% of students who graduated from high school in 2012 enrolled in college. Twenty n

Usimov [2.4K]

Answer:

a) The probability that exactly 17 of them enroll in college is 0.116.

b) The probability that more than 14 enroll in college is 0.995.

c) The probability that fewer than 11 enroll in college is 0.001.

d) It would be be unusual if more than 24 of them enroll in college since the probability is 0.009.

Step-by-step explanation:

We can model this with a binomial distribution, with n=29 and p=0.65.

The probability that k students from the sample who graduated from high school in 2012 enrolled in college is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{29}{k} 0.65^{k} 0.35^{29-k}\\\\\\$

a) The probability that exactly 17 of them enroll in college is:

$P(x=17) = \dbinom{29}{17} p^{17}(1-p)^{12}=51895935*0.0007*0=0.116\\\\\\$

b) The probability that more than 14 of them enroll in college is:

$P(X>14)=\sum_{15}^{29} P(X=k_i)=1-\sum_{0}^{14} P(X=k_i)\\\\\\P(x=0)=0\\\\P(x=1)=0\\\\P(x=2)=0\\\\P(x=3)=0\\\\P(x=4)=0\\\\P(x=5)=0\\\\P(x=6)=0\\\\P(x=7)=0\\\\P(x=8)=0\\\\P(x=9)=0\\\\P(x=10)=0.001\\\\P(x=11)=0.002\\\\P(x=12)=0.005\\\\P(x=13)=0.013\\\\P(x=14)=0.027\\\\\\P(X>14)=1-0.005=0.995$

c) Using the probabilities calculated in the point b, we have:

$P(X$

d) The probabilities that more than 24 enroll in college is:

$P(X>24)=\sum_{25}^{29}P(X=k_i)\\\\\\ P(x=25) = \dbinom{29}{25} p^{25}(1-p)^{4}=23751*0*0.015=0.007\\\\\\P(x=26) = \dbinom{29}{26} p^{26}(1-p)^{3}=3654*0*0.043=0.002\\\\\\P(x=27) = \dbinom{29}{27} p^{27}(1-p)^{2}=406*0*0.123=0\\\\\\P(x=28) = \dbinom{29}{28} p^{28}(1-p)^{1}=29*0*0.35=0\\\\\\P(x=29) = \dbinom{29}{29} p^{29}(1-p)^{0}=1*0*1=0\\\\\\\\P(X>24)=0.007+0.002+0+0+0=0.009$