All categories

3 years ago

There are 4 cups in a quart and 4 quarts in a ganon.

Mathematics

2 answers:

larisa86 [58]3 years ago

7 0

Answer:

The answer is 80 cups of water in 5 gallons.

Step-by-step explanation:

Because there are 4 cups of water in one quart, you have to multiply four by four(four cups of water in a quart, and 4 quarts of water in a gallon) giving you 16 cups of water in 1 gallon. Then to figure out how many cups of water are in 5 gallons, you have to multiply 16 by 5 to get 80 cups of water in 5 gallons.

Assoli18 [71]3 years ago

6 0

Answer:20 quarts

Step-by-step explanation: if there are 4 quarts in a gallon and you need to know how many quarts are in five gallons just multiply 5 and 4 and you will get 20 quarts.

You might be interested in

Brody earns $14 per hour for the first 40 hours worked in a week and $21 per hour for any hours over 40 that he worked. Brody ea

laiz [17]

Answer:

$29,120

When you're earning $14 an hour, you'll be making $560 a week, $2,427 a month, and annually $29,120. This is because 14 x 40 x 52 makes 29,120

3 0

2 years ago

Read 2 more answers

Could someone solve this for me?

Alja [10]

you miss something, maybe log is lg.

3 0

3 years ago

Let f(t) represent the temperature of a turkey baking in an oven as a function of time (t) in the oven (in minutes). This means

WITCHER [35]

Yes yup yes yes we are just okay

5 0

3 years ago

What is 7/56 in simplest form?

nika2105 [10]

8. 56 divided by 7 is 8, the simplest form of it

8 0

4 years ago

Read 2 more answers

A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equ

Over [174]

Answer: The correct number of balls is (b) 4.

Step-by-step explanation: Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.

Let 'r' represents the number of balls to be picked up.

Since we are choosing from 10 balls, so we must have

$^{10}C_r=210.$

The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

$^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10$

If r = 2, then

$^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45$

If r = 3, then

$^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120$

If r = 4, then

$^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.$

Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.

Thus, (b) is the correct option.

4 0

4 years ago

Read 2 more answers