All categories

3 years ago

If a boy has 4 pairs of shorts, 8 shirts, and 2 pairs of shoes, how many different outfits could he wear?

Mathematics

1 answer:

sveta [45]3 years ago

4 0

Answer:

Step-by-step explanation:

I paired the first shirt with all of the pants and both shoes which was 8 outfits, but here are 8 other shirts which will also make 8 outfits.

You might be interested in

A local gym instructor has a course load that allows her to teach eight classes. At an interest meeting, 8 people wanted high-‐

cricket20 [7]

Answer:

The final curse load of the local gym instructor is this:

High-‐impact aerobics = 1

Low-‐impact aerobics = 4

Jazzercise = 1

Step exercise = 2

Step-by-step explanation:

1. Let's apportion the class load using the Hamilton method

Number of classes the local gym instructor can teach = 8

Total number of students that want to take a class = 114 (8 people wanted high-‐impact aerobics, 64 wanted low-‐impact aerobics, 11 wanted jazzercise, and 31 wanted step exercise)

Standard divisor = 114/8 = 14.25

Now, we can apportion the students in in Standard quotas, this way:

High-‐impact aerobics = 8/14.25 = 0.5614

Low-‐impact aerobics = 64/14.25 = 4.49

Jazzercise = 11/14.25 = 0.7719

Step exercise = 31/14.25 = 2.1754

Now, we find the Minimum quota, just considering the whole number and don't taking into account the decimals, this way:

High-‐impact aerobics = 0

Low-‐impact aerobics = 4

Jazzercise = 0

Step exercise = 2

As we can see we have 6 classes and there are 2 still pending. Those 2 goes to the classes with the highest decimal portion, in this case, Jazzercise .7719 and High-‐impact aerobics with .5614.

The final course load is this:

High-‐impact aerobics = 1

Low-‐impact aerobics = 4

Jazzercise = 1

Step exercise = 2

7 0

3 years ago

Find the LCD 4/7 and 3/5

blsea [12.9K]

I would multiply 4/7 by 5 and 3/5 by 7 to get 20/35 and 21/35. The Least Common factor is 35

8 0

3 years ago

Suppose that the proportions of blood phenotypes in a particular population are as follows: A B AB O 0.48 0.13 0.03 0.36 Assumin

Serggg [28]

Answer:

P(O and O) =0.1296

P=0.3778

Step-by-step explanation:

Given that

blood phenotypes in a particular population

A=0.48

B=0.13

AB=0.03

O=0.36

As we know that when A and B both are independent that

P(A and B)= P(A) X P(B)

The probability that both phenotypes O are in independent:

P(O and O)= P(O) X P(O)

P(O and O)= 0.36 X 0.36 =0.1296

P(O and O) =0.1296

The probability that the phenotypes of two randomly selected individuals match:

Here four case are possible

P=P(A and A)+P(B and B)+P(AB and AB)+P(O and O)

P=0.48 x 0.48 + 0.13 x 0.13 + 0.03 x 0.03 + 0.36 x 0.36

P=0.3778

7 0

3 years ago

Can someone help me plz

Musya8 [376]

Answer:

Step-by-step explanation:

lets say the cost for a pound is 2$

if the watermelon weighs 3 pounds it would cost $6.

2(3)

The weight is the variable in the parenthesis because that is what is constantly changing.

4 0

3 years ago

PLZ HELP I GIVE BRAINLIEST!!!

GREYUIT [131]

Answer:

800 × ( 3.87÷ 100 ) × 1

= 30.96

8 0

3 years ago