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Ludmilka [50]
3 years ago
6

A company produces fruity drinks that contain a percentage of real fruit juice. Drink A contains 20% real fruit juice and Drink

B contains 10% real fruit juice. How many liters of real fruit juice would be needed to produce 250 liters of Drink A and 150 liters of Drink B? How many liters of real fruit juice would be needed to produce aa liters of Drink A and bb liters of Drink B?
Mathematics
1 answer:
sweet [91]3 years ago
4 0

Answer:

y = 290 liters Drink B

Step-by-step explanation:

Drink A contains 20% real fruit juice. x liters

and Drink B contains 5% real fruit juice. y liters

y=x+80

20%x +5%y =56.5

20x+5y==5650

-x + y = 80.00

20x + 5 y = 5650.00 .............2

Eliminate y

multiply (1)by -5

Multiply (2) by 1

5.00 x -5.00 y = -400.00

20.00 x 5.00 y = 5650.00

Add the two equations

25.00 x = 5250.00

/ 25.00

x = 210.00

plug value of x in (1)

-1.00 x + 1.00 y = 80.00

-210.00 + 1.00 y = 80.00

1.00 y = 290.00

y = 290.00

Ans x = 210 liters Drink A

y = 290 liters Drink B

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Read 2 more answers
e chairman of the statistics department in a certain college believes that 70% of the department’s graduate assistantships are g
8_murik_8 [283]

Answer:

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

Step-by-step explanation:

Problems of normally distributed samples are 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.

For a proportion p in a sample of size n, we have that \mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

p = 0.7, n = 50

So

\mu = 0.7, \sigma = \sqrt{\frac{0.7*0.3}{50}} = 0.0648

What is the probability that the sample proportion will NOT be between 0.60 and 0.73?

This is 1 subtracted by the probability that it is between 0.6 and 0.73.

Probability it is between 0.6 and 0.73

pvalue of Z when X = 0.73 subtracted by the pvalue of Z when X = 0.6. So

X = 0.73

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

Z = \frac{0.73 - 0.7}{0.0648}

Z = 0.46

Z = 0.46 has a pvalue of 0.6772

X = 0.6

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

Z = \frac{0.6 - 0.7}{0.0648}

Z = -1.54

Z = -1.54 has a pvalue of 0.0618

0.6772 - 0.0618 = 0.6154

NOT be between 0.60 and 0.73?

1 - 0.6154 = 0.3846

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

5 0
4 years ago
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