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stellarik [79]
3 years ago
7

Put it into a scientific notation

Mathematics
2 answers:
cupoosta [38]3 years ago
5 0

Answer:

<h2>Hi! The answer to your question is 2.41860465 x 10^-2</h2>

Step-by-step explanation:

☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆

☁Brainliest is greatly appreciated!☁

<em>Hope this helps!!</em>

<em>- Brooklynn Deka</em>

Angelina_Jolie [31]3 years ago
3 0

Answer:

2.41860465*10^-2

Step-by-step explanation:

Just use your calculator

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-4 in the first and second box

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Step-by-step explanation:

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3 years ago
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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
Serga [27]

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:

\mu=P_g*n=0.5886*90=52.9746\\\\\sigma=\sqrt{np(1-p)}=\sqrt{90*0.5886*(1-0.5886)} =4.6684

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

To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 78 green balls in the sample:

0.65*120=78

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

z=\frac{x-\mu}{\sigma}=\frac{78-70.6320}{5.3905}=1.37

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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Step-by-step explanation:

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