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3 years ago

Find the distance between (-2,6) ,(4,8)

Mathematics

1 answer:

Yuri [45]3 years ago

5 0

Answer:

$\sqrt{40}$ OR 6.324555

Remember to round accordingly if not intended to be answered in square root form!

Step-by-step explanation:

Use the distance formula

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The law of Cosines Is......

iogann1982 [59]

Answer: $\Large\boxed{C.~21}$

Step-by-step explanation:

The goal of the question

Find the value of 2abcosC

Given information

a = 4

b = 3

c = 2

Given formula (Law of Cosine)

$c^2=a^2+b^2-2abcosC$

Substitute values into the formula (Leave 2abcosC as it is)

$(2)^2=(4)^2+(3)^2-2abcosC$

Simplify the exponents

$4=16+9-2abcosC$

Simplify by addition

$4=25-2abcosC$

Add 2abcosC on both sides

$4+2abcosC=25-2abcosC+2abcosC$

$4+2abcosC=25$

Subtract 4 on both sides

$4 + 2abcosC-4=25-4$

$\Large\boxed{2abcosC=21}$

Hope this helps!! :)

Please let me know if you have any questions

5 0

1 year 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

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 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 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

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

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$