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Elodia [21]
2 years ago
6

Mike's closing costs will add up to 4 percent and he'll make a down payment of 20 percent on a house that costs $210,000. Over t

he life of his loan, he will pay $197,040.76 in monthly payments. What is the total cost of his house? A. 244,540.76 B. 247,440.76 C. 265,440.76
Mathematics
1 answer:
postnew [5]2 years ago
7 0
<span>For the question "Mike's closing costs will add up to 4 percent and he'll make a down payment of 20 percent on a house that costs $210,000. Over the life of his loan, he will pay $197,040.76 in monthly payments. What is the total cost of his house?" To obtain the total cost of the house, we first obtain the amount he paid as the closing costs and the down payment he paid which we will add to the total amount paid in monthly payments. Closing cost = 4% of $210,000 = 0.04 x 210,000 = $8,400 Down payment = 20% of $210,000 = 0.2 x 210,000 = $42,000 Total monthly payments = $197,040.76 Total cost of the house = $8,400 + $42,000 + $197,040.76 = $247,440.76</span>
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8 0
2 years ago
A bicycle is traveling at 20 feet per minute. What is the bicycle's speed expressed in inches per second?
Vikentia [17]

Answer:

4 inches per second

Step-by-step explanation:

Speed = distance/time

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7 0
2 years ago
A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
6 0
2 years ago
Read 2 more answers
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1 year ago
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2 years ago
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