Please help with these two, or one I will give brainly for correct answer no bots or links

Evaluate the given expression if x = 25, y = 10, w = 11, and z = 18. (see the image)

hammer [34]

Answer:

D

Step-by-step explanation:

So we have the expression:

$(x-y)^2+10wz$

And we want to evaluate it when x=25, y=10, w=11, and z=18.

So, substitute these values for the variables:

$=(25-10)^2+10(11)(18)$

First, subtract within the parentheses:

$=(15)^2+10(11)(18)$

Square 15:

$=225+10(11)(18)$

Now, multiply the terms on the right. 10 times 11 is 110:

$=225+110(18)$

Multiply:

$=225+1980$

Finally, add:

$=2205$

So, our answer is D.

And we're done!

8 0

3 years ago

Read 2 more answers

160 bananas;20%decrease

Jobisdone [24]

Answer:

128 banana's... I would work it all out for u but the teacher just came in my room and had a go

7 0

2 years ago

Read 2 more answers

What is an equation of the line that passes through the point ( 6 , − 2 ) (6,−2) and is perpendicular to the line 6 x + y = 2 6x

Diano4ka-milaya [45]

Answer:

An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

$y=\frac{1}{6}x-3$

Step-by-step explanation:

We know that the slope-intercept form of the line equation is

y=mx+b

where m is the slope and b is the y-intercept.

Given the line

6x+y=2

Simplifying the equation to write into the slope-intercept form

y = -6x+2

So, the slope = -6

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.

Thus, the slope of the perpendicular line will be: -1/-6 = 1/6

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be

$y-y_1=m\left(x-x_1\right)$

substituting the values m = 1/6 and the point (6, -2)

$y-\left(-2\right)=\frac{1}{6}\left(x-6\right)$

$y+2=\frac{1}{6}\left(x-6\right)$

subtract 2 from both sides

$y+2-2=\frac{1}{6}\left(x-6\right)-2$

$y=\frac{1}{6}x-3$

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

$y=\frac{1}{6}x-3$

5 0

3 years ago

A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose pe

soldier1979 [14.2K]

Using the normal distribution, it is found that:

a) 0.8599 = 85.99% probability that x is more than 60.

b) 0.1788 = 17.88% probability that x is less than 110.

c) 0.6811 = 68.11% probability that x is between 60 and 110.

d) 0.0643 = 6.43% probability that x is greater than 125.

In a normal distribution 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.

In this problem:

The mean is of 87, thus $\mu = 87$ .
The standard deviation is of 25, thus $\sigma = 25$ .

Item a:

This probability is 1 subtracted by the p-value of Z when X = 60, thus:

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

$Z = \frac{60 - 87}{25}$

$Z = -1.08$

$Z = -1.08$ has a p-value of 0.1401.

1 - 0.1401 = 0.8599

0.8599 = 85.99% probability that x is more than 60.

Item b:

This probability is the p-value of Z when X = 110, thus:

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

$Z = \frac{110 - 87}{25}$

$Z = 0.92$

$Z = 0.92$ has a p-value of 0.8212.

1 - 0.8212 = 0.1788.

0.1788 = 17.88% probability that x is less than 110.

Item c:

This probability is the p-value of Z when X = 110 subtracted by the p-value of Z when X = 60.

From the previous two items, 0.8212 - 0.1401 = 0.6811.

0.6811 = 68.11% probability that x is between 60 and 110.

Item d:

This probability is 1 subtracted by the p-value of Z when X = 125, thus:

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

$Z = \frac{125 - 87}{25}$

$Z = 1.52$

$Z = 1.52$ has a p-value of 0.9357.

1 - 0.9357 = 0.0643.

0.0643 = 6.43% probability that x is greater than 125.

A similar problem is given at brainly.com/question/24863330

7 0

3 years ago

What is the approximate area of the shaded sector in the circle shown below?

stich3 [128]

Answer: 26.5cm2

Step-by-step explanation:

4 0

3 years ago