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Igoryamba
1 year ago
12

Two college roommates have each committed to donating to charity each week for the next year. The roommates’ weekly incomes are

independent of each other. Suppose the amount donated in a week by one roommate is approximately normal with mean $30 and standard deviation $10, and the amount donated in a week by the other roommate is approximately normal with mean $60 and standard deviation $20. Which of the following is closest to the expected number of weeks in a 52-week year that their combined donation will exceed $120 ?0; the combined donation never exceeds $120 in a weekA1 weekB3 weeksC5 weeksD8 weeks
Mathematics
1 answer:
schepotkina [342]1 year ago
6 0

Answer:

C. 5 weeks.

Step-by-step explanation:

In this question we have a random variable that is equal to the sum of two normal-distributed random variables.

If we have two random variables X and Y, both normally distributed, the sum will have this properties:

S=X+Y\\\\\ \mu_S=\mu_X+\mu_Y=30+60=90\\\\\sigma_S=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}=22.36

To calculate the expected weeks that the donation exceeds $120, first we can calculate the probability of S>120:

z=\frac{S-\mu_S}{\sigma_S} =\frac{120-90}{22.36}=\frac{30}{22.36}=1.34\\\\P(S>120)=P(z>1.34)=0.09012

The expected weeks can be calculated as the product of the number of weeks in the year (52) and this probability:

E=\#weeks*P(S>120)=52*0.09012=4.68

The nearest answer is C. 5 weeks.

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A
gavmur [86]

Answer:

m∠QPM=43°

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠NPQ=m∠MPN+m∠MPQ

we have

m∠NPQ=(9x-25)°

m∠MPN=(4x+12)°

m∠MPQ=(3x-5)°

substitute the given values and solve for x

(9x-25)°=(4x+12)°+(3x-5)°

(9x-25)°=(7x+7)°

9x-7x=25+7

2x=32

x=16

Find the measure of angle QPM

Remember that

m∠QPM=m∠MPQ

m∠MPQ=(3x-5)°

substitute the value of x

m∠MPQ=(3(16)-5)=43°

therefore

m∠QPM=43°

7 0
2 years ago
Which equation represents a line that passes through (4, left-parenthesis 4, StartFraction one-third EndFraction right-parenthes
lina2011 [118]

Answer:

y-\frac{1}{3}=\frac{3}{4}(x-4)              

Step-by-step explanation:              

we know that            

The equation of the line into point slope form is equal to                        

y-y1=m(x-x1)              

In this problem we have          

(x_1,y_1)=(4,\frac{1}{3})                            

m=\frac{3}{4}            

substitute the given values                

y-\frac{1}{3}=\frac{3}{4}(x-4)

therefore

y minus StartFraction one-third EndFraction equals StartFraction 3 Over 4 EndFraction left-parenthesis x minus 4 right-parenthesis.(x – 4)

             

6 0
2 years ago
Read 2 more answers
The table below shows the price of rolls at a bakery. A baker’s dozen includes 13 rolls for the price of a dozen rolls. How much
Tpy6a [65]
Please provide table or else i can't calculate!
Answer: is ) $0.60 cents.
3 0
2 years ago
Read 2 more answers
Recent homebuyers from a local developer allege that 30% of the houses this developer constructs have some major defect that wil
umka21 [38]

Answer:

3.54% probability of observing at most two defective homes out of a random sample of 20

Step-by-step explanation:

For each house that this developer constructs, there are only two possible outcomes. Either there are some major defect that will require substantial repairs, or there is not. The probability of a house having some major defect that will require substantial repairs is independent of other houses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

30% of the houses this developer constructs have some major defect that will require substantial repairs.

This means that p = 0.3

If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20

This is P(X \leq 2) when n = 20. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.3)^{0}.(0.7)^{20} = 0.0008

P(X = 1) = C_{20,1}.(0.3)^{1}.(0.7)^{19} = 0.0068

P(X = 2) = C_{20,2}.(0.3)^{2}.(0.7)^{18} = 0.0278

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0008 + 0.0068 + 0.0278 = 0.0354

3.54% probability of observing at most two defective homes out of a random sample of 20

5 0
1 year ago
Find the volume of a right circular cone that has a height of 12.2 cm and a base with a circumference of 18.5 cm. Round your ans
loris [4]

Answer:

110.7 cm³ to the nearest tenth

Step-by-step explanation:

The volume V of a cone whose circular base has a radius r and a height h is given as

V = 1/3 πr²h

where π = 22/7

Given the circumference of the circular base as 18.5cm, we must find the radius before we can computed the volume. The relationship between the radius and the circumference is such that the circumference C

= 2πr

hence,

r = C/2π

= 18.5/2π

since the height of the cone is 12.2 cm, the volume V

= 1/3 * π * (18.5/2π)² * 12.2

= 110.71 cm³

5 0
2 years ago
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