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2 years ago

Struggling with homework!!

Mathematics

1 answer:

sveta [45]2 years ago

8 0

Answer:

Step-by-step explanation:

if 20% were Caesar and 20 divided by 5 is 4, then every four salads sold is five percent.

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Rochelle needs to pack a picture frame that measures 7611 inches on each side. Will it fit into a box that measures 7.52 inches

Ket [755]

Answer:

No, the picture frame will not fit.

7 6/11=7.545454...

7.545454... is larger than 7.5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Kayla purchased a used truck that had 27,544 kilometers on it. Later on, she sold the truck when it had 80,452 kilometers on it.

Ad libitum [116K]

Answer:

Kayla put 52,908 km on the truck.

Step-by-step explanation:

Subtract 27,544 km from 80,452 km to answer this question:

80,452 km

- 27,544 km

-------------------

52,908 km

Kayla put 52,908 km on the truck.

3 0

2 years ago

Tell me the answer of this

Harrizon [31]

The answer you are looking for is: 
 $\frac{3}{2}$ 
hope that helps!! 
Have a wonderful day!!

4 0

3 years ago

494 is 152% of what number?

Darya [45]

Answer:

747.84

Step-by-step explanation:

change the percent to a decimal 1.52 then multiply by 494

1.52×494=747.84

6 0

2 years ago

How would i solve a problem like this?

SVETLANKA909090 [29]

The number of combinations of size $k$ that you can make with $n$ items is given by the so-called binomial coefficient,

$\dbinom nk = \dfrac{n!}{k!(n-k)!}$

• $n!$ is the number of ways of permuting $n$ items.

• $(n-k)!$ is the number of ways of permuting all but $k$ of the $n$ items.

Dividing $n!$ by $(n-k)!$ then gives the number of ways of permuting only $k$ of the total $n$ items.

• $k!$ is the number of ways of permuting $k$ items.

Dividing $\frac{n!}{(n-k)!}$ by $k!$ then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

$\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20$

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

$\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21$

was of selecting any 2 boys from the total 7 boys.

Then there are

$\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}$

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

$\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287$

Then the probability of selecting such a committee at random is

$\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263$

8 0

1 year ago