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yawa3891 [41]
3 years ago
15

What’s the recipical of 7/9

Mathematics
2 answers:
Kryger [21]3 years ago
8 0

<u>Answer:</u>

The reciprocal of \frac { 7 } { 9 } is \frac { 9 } { 7 }.

<u>Step-by-step explanation:</u>

We are asked about the reciprocal of the given fraction \frac { 7 } { 9 }.

We know that in mathematics, a reciprocal is the multiplicative inverse of a number.

For example, in case of a fraction, its reciprocal will be its inverse which means that we swap the places of numerator and denominator.

So the reciprocal of \frac { 7 } { 9 } will \frac { 9 } {7 }.

NikAS [45]3 years ago
5 0

Fractions are always set up with numerator on top and denominator on bottom like so:

\frac{numerator}{denominator}

To take the reciprocal of a number you must switch the places of numerator and denominator.

Reciprocal of \frac{7}{9} is \frac{9}{7}

Hope this helped!

~Just a girl in love with Shawn Mendes

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Answer:

the answer is 1/12

Step-by-step explanation:

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A golf ball is hit in not the air represented by the equation y= -3x^2+18x+45. The maximum height the ball will reach is____ fee
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Answer:

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Here, we want to get the maximum height the ball will reach

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By doing this, we have it that the vertex is at the point (3,72)

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Which expression is equivalent to 3(m - 3) + 4?
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Answer:

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2 years ago
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The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 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 = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

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