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3 years ago

I need help with this as well ty lol

Mathematics

2 answers:

Pie3 years ago

4 0

so you use the formula a^2 + b^2 = c^2
so C will be JL
13m^2 + 17m^2 = c^2
458 m = c^2

to leave C alone you need to square root it

√458m = √c^2
21.4 = C

Lady bird [3.3K]3 years ago

4 0

Answer:

about 21.4

following Pythagorean theorem

JK and KL are the legs

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Which of the following is the simplest form of this expression?

Readme [11.4K]

The numerator could be written a⁴/⁵

The denominator could be written a²/³

Now solve ( a⁴/⁵) / (a²/³) ==> a⁽⁴/⁵ ⁻²/³) = a⁽²/¹⁵)

This is the simplest way

4 0

3 years ago

Read 2 more answers

The triangles are similar.

MrMuchimi

25/5 = 5
and also
15/3 = 5
so
(3x+ 8)/5 = 4
3x + 8 = 20
3x = 12
x = 12/3
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answer is x = 4

7 0

3 years ago

Read 2 more answers

Plz urgent <br>inequalities

solmaris [256]

Answer:

X>-4

Y>-18

X>-7

I hope this helps

3 0

2 years ago

The property tax on a house with an assessed value of $624,000 is $7280. Determine the property tax on a house with an assessed

Law Incorporation [45]

Using proportions, the property tax on a house with an assessed value of $762,000 is of $8,890.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The tax rate, as a proportion, is:

7280/624000 = 0.011667.

Hence, the property tax on a house with an assessed value of $762,000 is of:

0.011667 x 762,000 = $8,890.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

3 0

1 year ago

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.

5 0

3 years ago