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bazaltina [42]
3 years ago
12

The Yoga class you wish to take is offering two price quotes: $108 for 5 weeks as (15 classes) or $125 for 8 weeks (24 classes).

How much per class would you save by taking the 8 week class?
A $2.10
B $2.05
C $1.99
D $1.89
Mathematics
2 answers:
WARRIOR [948]3 years ago
8 0

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Cost for 5 weeks i.e. 15 classes = $108

Cost of per class is given by

\frac{108}{15}=7.2

Cost for 8 weeks i.e. 24 classes = $125

Cost of per class is given by

\frac{125}{24}=5.21

Amount of saving of per class by taking the 8 week class is given by

7.2-5.21=\$1.99

Hence, Option 'C' is correct.

AURORKA [14]3 years ago
4 0
The answer is C)
108/15 - 125/24 = 1.99
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An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed fro
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Answer:

The probability is 70% that the sample mean amount of juice will be contained between 4.9168 ounces and 5.0832 ounces.

Step-by-step explanation:

To solve this question, the Normal probability distribution and the Central Limit Theorem are important.

Normal probability distribution

Problems of normally distributed samples 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 5, \sigma = 0.4, n = 25, s = \frac{0.4}{\sqrt{25}} = 0.08

The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

The lower end of this interval is the value of X when Z has a pvalue of 0.5 - 0.7/2 = 0.15

The upper end of this interval is the value of X when Z has a pvalue of 0.5 + 0.7/2 = 0.85

Lower end

X when Z has a pvalue of 0.15. So X when Z = -1.04.

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

Due to the Central Limit Theorem

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

-1.04 = \frac{X - 5}{0.08}

X - 5 = -1.04*0.08

X = 4.9168

Upper end

X when Z has a pvalue of 0.15. So X when Z = 1.04.

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

Due to the Central Limit Theorem

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

1.04 = \frac{X - 5}{0.08}

X - 5 = 1.04*0.08

X = 5.0832

The probability is 70% that the sample mean amount of juice will be contained between 4.9168 ounces and 5.0832 ounces.

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