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3 years ago

Pleas help it’s due tomrrow

Mathematics

1 answer:

katovenus [111]3 years ago

5 0

Answer:

Step-by-step explanation:

K is located at -25

L is located at -15

just move 5 units away from J on both sides

hope this helps <3

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Anna35 [415]

Answer:

Step-by-step explanation:

80-37 = 43

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2 years ago

Find the length of one edge of the cube. <br> Edge length = Answer inches

ycow [4]

Volume

V = l³

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3 years ago

Regular hexagon ABCDEF is shown below. Find the measure of angle x.

Aleks [24]

Answer:

ANSWER: 30

ANGLE BCD IS MADE UP OF 180 DEGREES ANGLE C IS 120 AND SINCE THE ENTIRE ANGLE IS 180 AND C AND D IS CONGRUENT AND 60 DEGREES IS LEFT WE DIVIDE BY 2 TO FIND OUR ANSWER OF 3O

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3 years ago

Factor completely.

tatyana61 [14]

Answer:

A.) (7t³ + 2k^4)(7t³ - 2k^4)

Step-by-step explanation:

Factor the following:

49 t^6 - 4 k^8

49 t^6 - 4 k^8 = (7 t^3)^2 - (2 k^4)^2:

(7 t^3)^2 - (2 k^4)^2

Factor the difference of two squares. (7 t^3)^2 - (2 k^4)^2 = (7 t^3 - 2 k^4) (7 t^3 + 2 k^4):

Answer: (7 t^3 - 2 k^4) (7 t^3 + 2 k^4)

3 0

3 years ago

Read 2 more answers

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$