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3 years ago

5

We can think of 6 divided by 3/4 as how many 3/4’s are in 6?

1 answer:

Flura [38]3 years ago

7 0

It means you divide $6$ with $\frac{3}{4}$ .

When you divide a fraction, you reciprocate it meaning divide it by $1$ or switch the numerator and the denominator. The reciprocated one will be: $\frac{4}{3}$

Multiply 6 with 4/3 and you get: $8$

$\frac{3}{4}$ goes $8$ times in $6$ :D

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Stu Dent has 7 different math books, 5 different science books and 10 different engineering books. Find the number of ways to pi

zepelin [54]

Answer:

155

Step-by-step explanation:

Number of math books = 7

Number of science books = 5

Number of engineering books = 10

We need to find the number of ways to pick two books with different subjects.

Solution :

We will use combination here .

If we need to choose r objects from total n objects ,

$n_{C_{r}}=\frac{n!}{r!(n-r)!}$

So, number of ways to pick two books of same subjects = $7_{C_{2}}+5_{C_{2}}+10_{C_{2}}=\frac{7!}{2!5!}+\frac{5!}{2!3!}+\frac{10!}{2!8!}\\\\=\frac{7\times 6}{2}+\frac{5\times 4}{2}+\frac{10\times 9}{2}\\\\=21+10+45=76$

Also, number of ways to select any two books = $22_{C_{2}}=\frac{22!}{2!20!}=\frac{22\times 21}{2}=21\times 11=231$

Therefore , number of ways to pick two books with different subjects =

number of ways to select any two books - number of ways to pick two books of same subjects = 231 - 76 = 155

3 0

3 years ago

Elina [12.6K]

10n=2.22
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4 years ago

Complete the statement, then write the postulate you used.

NemiM [27]

The answer is MN with a bar/line on top of it

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3 years ago

Assuming boys and girls are equally likely, find the probability of a couple having a baby girl when their sixth child is born

Ivanshal [37]

Answer: The required probability of having 6th girl is 0.5.

Step-by-step explanation: Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B} and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.$

Thus, the required probability of having 6th girl is 0.5.

3 0

3 years ago

Write -7 - (-4) as an addition problem. Then simplify the expression.

SVETLANKA909090 [29]

-7 + 4

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8 0

4 years ago