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Artemon [7]
4 years ago
7

Fruit-O-Rama sells dried pineapple for 0.25 per ounce. Mags spent a total of 2.65 on pineapple. How much ounces did she buy?

Mathematics
2 answers:
omeli [17]4 years ago
3 0
10 i think or 10.6 is the answer
nalin [4]4 years ago
3 0

Answer

Find out the how much ounces did she buy .

To proof

let us assume that the number of ounces did she buy be x .

As given

Fruit-O-Rama sells dried pineapple for $0.25 per ounce i.e the price of each pineapple is $0.25 ounce .

Mags spent a total of $2.65 on pineapple

than the equation becomes

x × 0.25 = 2.65

x = \frac{2.65}{0.25}

x = 10 .6 ounce

therefore  10 .6 ounce she buys .

Hence proved

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3 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
Help me on this question, I'll give brainlist for the correct answer.
german

Answer:

Step-by-step explanation:

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The second expression has value -6

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And finally the forth expression has value 1.14

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16/4 = 4

8 0
3 years ago
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