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harkovskaia [24]

3 years ago

5

There are 11 times as many cakes as there is pies. A worker counts that there are 60 more cakes than pies. How many pies is ther

e?

1 answer:

castortr0y [4]3 years ago

6 0

The number of pies present are 6

<u>Solution:</u>

Given that, there are 11 times as many cakes as there is pies.

A worker counts that there are 60 more cakes than pies.

We have to find that how many pies is there?

Let the number of pies be p, then number of cakes will be 11p.

Now, according to the given information,

<em>Number of cakes = 60 + number of pies </em>

11p = 60 + p

11p – p = 60

10p = 60

p = 6

Hence, the number of pies present are 6

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<u>Solution-</u>

From the figure,

AE = 2.4

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6 0

3 years ago

Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find

Papessa [141]

Answer:

1) $\text{P(at least one boy and one girl)}=\frac{3}{4}$

2) $\text{P(at least one boy and one girl)}=\frac{3}{8}$

3) $\text{P(at least two girls)}=\frac{1}{2}$

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To Find : The probability of each event.

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

1) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, BGG, GBB, GBG, GGB=6

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2) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, GBB=3

$\text{P(at least one boy and one girl)}=\frac{3}{8}$

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

$\text{P(at least two girls)}=\frac{4}{8}$

$\text{P(at least two girls)}=\frac{1}{2}$

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