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Black_prince [1.1K]
3 years ago
12

What is the present of 4 out of 30

Mathematics
1 answer:
alisha [4.7K]3 years ago
7 0

The present of 4 out of 30 is 13.33%

The present of 1 out of 30  is 3.33%

The present of 25 out of 30 is 83.33%

<u>Step-by-step explanation:</u>

  • The percentage is calculated by dividing the required number with the full quantity and multiplied by 100.

First sum :

(\frac{4}{30} )\times 100=13.33 \%

Second sum :

(\frac{1}{30} ) \times 100 = 3.33%

Third sum:

(\frac{25}{30} )\times100 = 83.33%

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Compute​ P(x) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate thi
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Answer:

The answers to the questions are;

A) P(X) where x = 12 is equal to 3.55 ×10⁻²

B) P(x) using the normal distribution is given by the probability distribution function and is equal to 3.453 ×10⁻²

C) The calculated probabilities of P(x=12) using the binomial probability formula and the probability distribution function differ by 9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     a. because np(1-p) %u2265 ⇒ np(1-p) ≥ 10

E) c. the value of fi represents the expected proportion of observation less than or equal to the ith data value.

Step-by-step explanation:

To solve the question, we note that

n = 58

p = 0.3

x = 12

Here we have n·p = 17.4 and n·q = 40.6 both ≥ 5 that is the normal distribution can be used to estimate the probability

A) We are to use the binomial probability formula to find P(X)

Where x = 12, we have P(12) = ₅₈C₁₂×0.3¹²×0.7⁴⁶

= 891794789340×0.000000531441×7.49×10⁻⁸ = 3.55 ×10⁻²

B) Using the standard distribution table we have

the z score for x = 12 given as z = \frac{x-\mu}{\sigma} = -1.55

From the normal distribution table, we have the probability that value is below 12 = .06057 while the normal probability distribution function which gives the probability of a number being 12 is given by

\frac{1}{\sigma\sqrt{2 \pi } } e^{\frac{-(x-\mu)^2}{2\sigma^2} }

where:

σ = Standard deviation = \sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999

μ = Sample mean = n·p = 58×0.3 =17.4

Therefore the probability density function is \frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}

C)  The probabilities differ by 3.55 ×10⁻² -  3.453 ×10⁻² =  9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     n·p and n·p·(n-p) ≥ 5

E)   f_i represents the area under the curve towards left of the ith data observed in a normally distributed population.

Therefore, the value of fi represents the expected proportion of observation less than or equal to the ith data value  

5 0
3 years ago
The tape diagram below represents 100% percent.
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The shaded area is 75%
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X^2-x-20 how do I write that in factored form
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Answer:

(x+4)(x-5)

Step-by-step explanation:

You need to find two numbers that multiply to be -20 but add up to be -1.

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12 is what percent of 24? What percent of 35 is 7? What percent of 90 is 9? 20 out of 60 is what percent?​
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Answer:

12 is 50% of 24

7 is 20% of 35

9 is 10% of 90

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work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
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Answer:

Step-by-step explanation:

Here is the complete question.

Dominique, Marco, Roberto , and John work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company decides that the two individuals who get to attend will have their names drawn from a hat. This is like obtaining a simple random sample of size 2. (a) Determine the sample space of the experiment. That is, list all possible simple random samples of size n = 2. (b) What is the probability that Dominique and Marco attend the conference? (c) What is the probability that John attends the conference?  ​(d) What is the probability that John stays​ home?

Since there are four employees to select the two to send from. then for us to get the sample space for the experiment we need to merge all possible two employees and represent them as set.

Let D = Dominique, M = Marco, R = Roberto and J = John

a) The sample space for the experiment is the total number of possible outcomes that we can have. It is as given below

S = 4C2 = 4!/(4-2)!2! (Selecting 2 out of 4 employees)

Total sample space = 4!/2!2!

Total sample space = 4*3*2!/2!2

Total sample space = 12/2 = 6

The sample space are S = {DM, DR, DJ, MR, MJ, RJ}

b) Probability is the ratio of number of event to the sample space.

P = n(E)/n(S)

Given n(S) = 6

n(E) is the event of Dominique and Marco attending the conference.

E = {DM}

n(E) = 1

P(D and M) = 1/6

Hence  the probability that Dominique and Marco attend the conference is 1/6

c) For John to attend the conference, the event outcome will be given as;

E = {DJ, MJ, RJ}

n(E) = 3

n(S) = 6

Probability for John to attend the conference is 3/6 = 1/2

d) Probability that John stays at home = 1 - Prob (John attends the conference)

Probability that John stays at home = 1 - 1/2

Probability that John stays at home = 1/2

7 0
3 years ago
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