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alekssr [168]
3 years ago
5

Subtract the fractions and simplify the answer 9- 5 1/3

Mathematics
2 answers:
pentagon [3]3 years ago
7 0
3 2/3
All you have to do is add my answer to 5 1/3 and if it is 9, then it is correct
oee [108]3 years ago
4 0
9 − 5 1/3

=9−1 6/3

=11/3

=3 2/3

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Express the following as a unit rate :<br><br> 5 croissants for $10
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Answer:

$2/croissant

Step-by-step explanation:

$10÷ 5croissants = $2/ 1 croissant

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Find the slope of the lines that passes through the points (1, -4) and (3, -1)​
lakkis [162]

Answer:

3/2

Step-by-step explanation:

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3 years ago
Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the
Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

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}}.

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.

In this problem, we have that:

\mu = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

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

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

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

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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4 0
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