What’s 6 divided by y minus 3 equals 15 divided by ?

I leave my family's vacation cabin at 8 a.m. and start driving home at a nice, safe 45 mph. Two hours later, my husband, who alw

Murljashka [212]

Answer:

2:30 pm

Step-by-step explanation:

In two hours (from 10am), she is 45 * 2 = 90 miles from vacation cabin.

From then on her average is still 45 mph and her husband's is 65 mph. So, husband catches her 20 miles (65-45=20) in every hour.

The husband has to catch up 90 miles. How long would it take if he is catching up at 20 miles an hour?

90/20 = 4.5 hours

So husband starts at 10am and 4.5 hours form that time is 2.30pm

6 0

4 years ago

A college committee consists of 3 freshmen, 4 sophomores, 5 juniors, and 2 seniors. A subcommittee of 4 consisting of 1 person f

dezoksy [38]

Answer:

120

Step-by-step explanation:

The subcommittee needs to contain 1 out of 3 freshmen, 1 out of 4 sophomores, 1 out of 5 juniors and 1 out of 2 seniors. Assuming order is not important, the number of possibilities to pick the subcommittee members is given by:

$N=5*4*3*2=120$

There are 120 possible subcommittees.

5 0

3 years ago

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it

gavmur [86]

Answer:

a) $P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b) $P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

c) A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

Purchased Gum Kept the Money Total

Students Given 4 Quarters 25 14 39

Students Given $1 Bill 15 29 44

Total 40 43 83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}$

And if we replace we got:

$P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}$

And if we replace we got:

$P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0

3 years ago

The picture below shows a right-triangle-shaped charging stand for a gaming system: The side face of a charging stand is a right

mamaluj [8]

Answer:

AB = 8/ cos 60

Step-by-step explanation:

We want to find the hypotenuse AB

Since we have a right triangle

cos theta = adj/ hyp

cos 60 = 8/ AB

AB cos 60 = 8

AB = 8/ cos 60

3 0

3 years ago

On thursday Bernies dog ate 1/4 lb more dog food than she ate on wednesday. On wednesday, the dog ate 1 1/3 times the amount of

kirill115 [55]

"On Thursday Bernie's dog ate 1/4 lb more dog food than she ate on Wednesday."

That means that on Wednesday, the dog ate 1/4 lb LESS that it ate on Thursday.

"On Wednesday, the dog ate 1 1/3 times the amount of food she ate on Tuesday."

That means that on Tuesday, the dog ate 1 1/3 times LESS that it ate on Wednesday.

Now start with Thursday: "If Bernie's dog ate 1 3/4 lb food on Thursday"

On Wednesday it ate 1/4 lb less that that, so on Wednesday it ate

1 3/4 - 1/4 = 1 2/4 = 1 1/2

On Wednesday it ate 1 1/2 lb.

On Tuesday it ate 1 1/3 times less than on Wednesday. Now we divide 1 1/2 by 1 1/3.

$= 1 \dfrac{1}{2} \div 1 \dfrac{1}{3}$

$= \dfrac{3}{2} \div \dfrac{4}{3}$

$= \dfrac{3}{2} \times \dfrac{3}{4}$

$= \dfrac{3 \times 3}{2 \times 4}$

$= \dfrac{9}{8}$

$= 1 \dfrac{1}{8}$

Answer: It ate 1 1/8 lb of food on Tuesday.

8 0

3 years ago