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Mathematics

1 answer:

trapecia [35]3 years ago

4 0

It is everyone of those except for (-6, 3).

Step-by-step explanation:

4 / 6 = 2/3

2/3 x -9 = -6

2/3 x -12 = -8

2/3 x 3 = 2

2/3 x -6 = -4

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3.) Alex makes $31,992 per year and pays about 15% of his gross monthly income in federal

Fiesta28 [93]

Answer: 1699.5 dollars

Step-by-step explanation:

If Alex makes $31,992 per year, then his monthly income will be

31992/12 = 2666 dollars

and he pays about 15% of his gross monthly income. That is

15/100 × 2666 = 399.9

The remaining money will be

2666 - 399.9 = 2266.1

He should able to afford 25% of his monthly income for rent

25/100 × 2266.1 = 567 dollars

The money left after tax and payment for rent will be

2266.1 - 567 = 1699.5 dollars

3 0

4 years ago

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:

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