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jek_recluse [69]
3 years ago
8

How to solve the problem

Mathematics
1 answer:
Nat2105 [25]3 years ago
7 0
The answer will be 1,260 because u have to multiply all the the numbers to get area
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Describe a time when you use math outside of school ( pls I need help with this writing prompt)
Serga [27]

Answer:

Counting my money

Step-by-step explanation:

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I will love you forever if someone can help me with this
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It should be B. right angle and C. complementary angles
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Suppose there is a 13.9 % probability that a randomly selected person aged 40 years or older is a jogger. In​ addition, there is
Blababa [14]

Answer:

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

It would be unusual to randomly select a person aged 40 years or older who is male and jogs.

Step-by-step explanation:

We have these following probabilities.

A 13.9% probability that a randomly selected person aged 40 years or older is a jogger, so P(A) = 0.13.

In​ addition, there is a 15.6% probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. I am going to say that P(B) is the probability that is a male. P(B/A) is the probability that the person is a male, given that he/she jogs. So P(B/A) = 0.156

The Bayes theorem states that:

P(B/A) = \frac{P(A \cap B)}{P(A)}

In which P(A \cap B) is the probability that the person does both thigs, so, in this problem, the probability that a randomly selected person aged 40 years or older is male and jogs.

So

P(A \cap B) = P(A).P(B/A) = 0.156*0.139 = 0.217

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

A probability is unusual when it is smaller than 5%.

So it would be unusual to randomly select a person aged 40 years or older who is male and jogs.

4 0
3 years ago
A metal bar weighs 7 Kg. 93% of the bar is Silver. How many Kgs of silver is there in the bar?​
Sindrei [870]
<h3>Solution:</h3>

<u>From the given data</u>

<u>93% of the bar is silver :</u>

\small\bold{ → 93\%    \: \: of \: \:   7kg}

\small \bold{→\frac{93}{100}  \times 7 }

\small \bold{→ \frac{651}{100}  }

\small \bold{→ 6.51 \: kg  }

<u>Therefore, 6.51 Kg of Silver is there in the bar.</u>

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2 years ago
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2 point something most likely. Byeeee
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