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Radda [10]
2 years ago
11

In a school there are 542 girls and boys. how many pupils are there in that school?

Mathematics
2 answers:
DaniilM [7]2 years ago
6 0

Answer:

There are 1084 pupils in that school.

Step-by-step explanation:

If there are 542 girls and boys then you would have to double 542 to get the total number of people. So, there are 1084 people in that school.

Hope this helps!

goldenfox [79]2 years ago
3 0

Answer:

1084

Step-by-step explanation:

542+542

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Fourteen and one eighth minus twelve and five ninths
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1 and twenty thirty sixths

Step-by-step explanation:

14 1/8 - 12 5/9

1/8x4/4=4/36

1/8= 4/36

5/9x4/4= 20/36

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3 years ago
Negate the conditional statement.<br><br> c⇒(a∧∼b)
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The negate of this conditional statement as : a∨(∼b).

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We have the the following conditional statement -

c ⇒ (a∧∼b)

We can write the negate of this conditional statement as -

a∨(∼b)

Therefore, the negate of this conditional statement as : a∨(∼b).

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Find the quotient.<br> 0.42<br> -0.7<br><br><br> HELP PLEASE
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6 0
3 years ago
Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case
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Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   =1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

Thus, the probability that neither Dave nor Mike will show up is 0.25.

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3 years ago
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