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MariettaO [177]

3 years ago

8

Find the value of K to complete the square for the quadratic function y=x to the second power - 6x + k PLEASE HELP IT WOULD BE A

PPRECIATED IF U HELPED MEE

1 answer:

melisa1 [442]3 years ago

7 0

It’s -k which will be 1 because you need to isolate it

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Can someone help me with this

gulaghasi [49]

Answer:

Y is 90 degrees.

Step-by-step explanation:

This is an equalterial triangle which means all of the angles are the same. The angle for the upper part of the triangle is 60 as well. We can do 2x=60 because they add up to the angle which is 60. This gives us x=30. If x=30 and the other angle is 60, then the other angle has to be 90 because all triangle has an angle sum of 180. So 30 +60 + 90= 180. Hope this helps!

4 0

3 years ago

Read 2 more answers

3. The adult men of the Dinaric Alps have the highest average height of all regions. The

ahrayia [7]

Using the normal distribution, it is found that:

3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

3 - b) The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

4 - a) The 25th percentile for the math scores was of 71.6 inches.

4 - b) The 75th percentile for the math scores was of 78.4 inches.

<h3>Normal Probability Distribution </h3>

In a <em>normal distribution </em>with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

Question 3:

The mean is of 73 inches, hence $\mu = 73$ .

The standard deviation is of 3 inches, hence $\sigma = 3$ .

Item a:

The 40th percentile is X when Z has a p-value of 0.4, so <u>X when Z = -0.253</u>.

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

$-0.253 = \frac{X - 73}{3}$

$X - 73 = -0.253(3)$

$X = 72.2$

The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

Item b:

The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so <u>X when Z = 1.28</u>.

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

$1.28 = \frac{X - 73}{3}$

$X - 73 = 1.28(3)$

$X = 76.84$

The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

Question 4:

The mean score is of 75, hence $\mu = 75$ .

The standard deviation is of 5, hence $\sigma = 5$ .

Item a:

The 25th percentile is X when Z has a p-value of 0.25, so <u>X when Z = -0.675</u>.

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

$-0.675 = \frac{X - 75}{5}$

$X - 75 = -0.675(5)$

$X = 71.6$

The 25th percentile for the math scores was of 71.6 inches.

Item b:

The 75th percentile is X when Z has a p-value of 0.25, so <u>X when Z = 0.675</u>.

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

$0.675 = \frac{X - 75}{5}$

$X - 75 = 0.675(5)$

$X = 78.4$

The 75th percentile for the math scores was of 78.4 inches.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213

5 0

2 years ago

Evgesh-ka [11]

The answer will end up being c

6 0

3 years ago

Thirteen percent of the profits from Kendall’s business are donated to charities.write and solve an equation to find out how muc

pychu [463]

The amount given to charity from Kendall’s profit is $168.0146

Given:

Kendall’s profit = $1292.42
Percentage given to charity = 13%

<h3>How to find percentage</h3>

Amount given to charity = Percentage given to charity × Kendall’s profit

= 13% of $1292.42

= 13/100 × 1292.42

= 0.13 × 1292.42

= $168.0146

Therefore, the amount given to charity from Kendall’s profit is $168.0146

Learn more about percentage:

brainly.com/question/843074

5 0

2 years ago

The amount of time needed to complete a certain road trip, t, varies inversely with the speed of a

JulsSmile [24]

Answer:

x = 60

Step-by-step explanation:

t = $\frac{k}{x}$

15 = k/68

k=1020

~~~~~~~~~~~

17 = 1020/x

x = 1020/17

x = 60

6 0

3 years ago