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3 years ago

What 4 numbers adding give you a sum of 15

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

5 0

9+1+3+2= 15

Hope this helps you:)

alekssr [168]3 years ago

4 0

Answer: 6 and 9, 7 and 8. 69, 78, 87, 96. So, there are only 4 combinations which would result in 15.

Step-by-step explanation:

Therefore the average sum of three numbers is 45:3=15. The number 15 is called the magic number of the 3x3 square. You can also achieve 15, if you add the middle number 5 three times. The odd numbers 1,3,7, and 9 occur twice in the reductions, the even numbers 2,4,6,8 three times and the number 5 once.

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Read 2 more answers

At a new exhibit in the museum of science, people are asked to choose between 90 or 120 random draws from a machine. The machine

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Answer:

There is a probability of P₉₀=0.09853 of winning the prize with a 90-balls sample.

There is a probability of P₁₂₀=0.0853 of winning the prize with a 120-balls sample.

Step-by-step explanation:

In this problem, the balls that are removed are then replaced in the total, so the probability of obtaining a green ball is constant and is equal to:

$P_g=\frac{93}{93+65}= 0.5886$

This type of sampling should be analyzed with the binomial distribution, but given the sample size, it is convenient to use the approximation to a normal distribution.

The parameters of the normal distribution are:

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To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 59 green balls in the sample:

$0.65*90=58.5$

To calculate the probability of getting 59 or more green balls, we have to calculate the z-value

$z=\frac{x-\mu}{\sigma}=\frac{59-52.9746}{4.6684}= 1.29$

Then, the probability can be look up with the z-value in a normal distribution table:

$P(X \geq 59)=P(z\geq 1.29)=0.09853$

There is a probability of P=0.09853 of winning the prize with a 90-balls sample.

When the sample is of 120 balls, we have to recalculate the parameters.

Normal distribution

$\mu=P_g*n=0.5886*120=70.6320\\\\\sigma=\sqrt{np(1-p)}=\sqrt{120*0.5886*(1-0.5886)} =5.3905$

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$0.65*120=78$

To calculate the probability of getting 78 or more green balls, we have to calculate the z-value

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Then, the probability can be look up with the z-value in a normal distribution table:

$P(X \geq 78 )=P(z\geq 1.37)=0.0853$

There is a probability of P=0.0853 of winning the prize with a 120-balls sample.

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