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liubo4ka [24]
3 years ago
12

What number can 40 over 80 can be simplified to

Mathematics
2 answers:
Dmitrij [34]3 years ago
5 0
40/80 can be simplified to 1/2, by dividing 40 top and bottom

1/2 is your answer

hope this helps
ohaa [14]3 years ago
3 0
2 because 40 goes into 80 twice.


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4 adults consumed food costing $60 for 3days. For the same food cost ,what would be the cost of food consumed by 7 adults for 5d
salantis [7]

Answer:

$175

Step-by-step explanation:

$60÷3÷4 =5

$5 is the cost of one adult's food for one day.

$5×7×5

=$175

$175 is the cost of food consumed by 7 adults for 5 days.

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How many ounces in 2 pints
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32 US Fluid Ounces

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natulia [17]

Answer:

(A) 21

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The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

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.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
2 years ago
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