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motikmotik
3 years ago
9

There are 396 marbles in a bag. The bag contains red, yellow, and purple marbles. If the ratio of red to yellow marbles is 2:3 a

nd the ratio of yellow to purple marbles is 1:2, how many yellow marbles are in the bag?
Mathematics
1 answer:
Aneli [31]3 years ago
8 0

The no. of Red, Yellow and Purple marbles are 72, 108, 216 respectively.

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

Let us consider the no. of red marbles be R

Let the no. of yellow marbles be Y

Let the no. of purple marbles be P

Given that there are 396 marbles in the bag.

Total marbles = R+ Y + P

   396 = R + Y + P

The ratios are given as below:

   R : Y =2/3

Which means,  R/Y = 2/3

       R= 2Y/3

       Y: P = 1/2

Which means, Y/P = 1/2

       1/P = 1/2Y

         P = 2Y

Here we got the values of P and R in terms of Y.

Total marbles = R + Y + P

   396 =( 2Y/3) +Y+(2Y)

Taking LCM, we have 3 as LCM

   396 = (2Y +3Y+ 6Y) / 3

   396 x 3 = 2Y + 3Y + 6Y

   1188 = 11Y

   11Y = 1188

   Y = 1188/11

   Y = 108

R/Y = 2/3

R/108 = 2/3

R = 2(108) /3

R= 72

Y/P = 1/2

108/ P = 1/2

1/P = 1/2(108)

P = 2(108)

P = 216

The no. of Red, Yellow and Purple marbles are 72, 108, 216 respectively.

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3 years ago
Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 104 eligible vot
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Answer:

The probability that exactly 27 of 104 eligible voters voted is​ 0.057 = 5.7%.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this case, assume that 104 eligible voters aged 18-24 are randomly selected.

This means that n = 104.

Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

This means that p = 0.22

Mean and standard deviation:

\mu = 104*0.22 = 22.88

\sigma = \sqrt{104*0.22*0.78} = 4.2245

Probability that exactly 27 voted

By continuity continuity, 27 consists of values between 26.5 and 27.5, which means that this probability is the p-value of Z when X = 27.5 subtracted by the p-value of Z when X = 26.5.

X = 27.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{27.5 - 22.88}{4.2245}

Z = 1.09

Z = 1.09 has a p-value of 0.8621

X = 26.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{26.5 - 22.88}{4.2245}

Z = 0.86

Z = 0.86 has a p-value of 0.8051

0.8621 - 0.8051 = 0.057

The probability that exactly 27 of 104 eligible voters voted is​ 0.057 = 5.7%.

5 0
3 years ago
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