8x + 3y = 4<br> - 7x + 5y = -34

The daily low temperatures last week are shown below. What is the mean

kirill115 [55]

Answer:

i do not see the tempatures but add them all up and divide by however many tempatures there are

Step-by-step explanation:

5 0

3 years ago

*Reminder: If you have not filled out your graphic organizer to reflect the order of events in Part I of "The Monkey's

Andrew [12]

The two events that created the most tension in the story include:

Mrs. White suggests that her husband wishes for four hands for her.
Mr. White paid Sergeant-Major Morris a trifle for the paw.

<h3>What is tension?</h3>

It should be noted that tension in literature simply means the sense of something ominous in the literary work. It's a feeling that something is about to happen.

The Monkey's Paw is a horror short story by author W. W. Jacobs. In the story, three wishes are granted to the owner of The Monkey's Paw, but the wishes come with a price for interfering with fate.

It should be noted that Mrs. White suggesting that her husband wish for four hands for her created tension in the story as the readers were eager to know what happened in the story. Also, it should be noted Mr. White paying Sergeant-Major Morris a trifle for the paw created tension.

Learn more about tension on:

brainly.com/question/918617

#SPJ1

8 0

1 year ago

What is the value of x?

dangina [55]

Answer:

Step-by-step explanation:

The angles add up to 360°.

x + 98 + x + 91 + x + 15 = 360
3x + 204 = 360
3x = 156
<u>x = 52°</u>

7 0

2 years ago

What is the area of the fire pit? (Use 3.14 for pi.)

laila [671]

Answer:

$Area = 65.94$

Step-by-step explanation:

I will assume that the shaded region is the fire pit

Given

$R = 5$

$r = 2$

Required

Determine the area of the fire pit

First, calculate the area of the big circle.

$A_2 = \pi R^2$

Area of the small is:

$A_1 = \pi r^2$

The area of the pit is:

$Area = A_2 - A_1$

$Area = \pi R^2 - \pi r^2$

$Area = \pi (R^2 - r^2)$

$Area = 3.14 (5^2 - 2^2)$

$Area = 3.14 (25 - 4)$

$Area = 3.14 * 21$

$Area = 65.94$

5 0

3 years ago

For this problem, assume that three students are chosen from 5 freshmen, 4 sophomores, and 4 juniors.

Sati [7]

The formula $C(n, r)= \frac{n!}{r!(n-r)!}$ , where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

<span>The sample space S in our problem is as follows: </span>

triples selected from 5 freshmen, 4 sophomores, and 4 juniors, so that at least 1 freshman is selected.

The selection of 1 freshman can be done in 5 ways. The selection of the remaining 2 out of 4 freshmen (one is already chosen), 4 sophomores, and 4 juniors can be done in C(12, 2) many ways.

$C(12, 2)=\displaystyle{ \frac{12!}{2!10!}= \frac{12\cdot11\cdot10!}{2!10!}=66$

So there are in total, 5*66= 330 possible triples.

n(S)=330

The event E "selecting 2 freshmen and 1 junior" can occur in :

$C(5, 2)\cdot 4=\displaystyle{ \frac{5!}{2!3!}\cdot4=40$ many ways.

so n(E)=40

P(E)=n(E)/n(S)=40/330=0.12

Answer: 0.12

4 0

4 years ago

8x + 3y = 4 - 7x + 5y = -34