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Step2247 [10]
2 years ago
8

Yo plz help me I don’t know this Which number line shows the solution to -6 - (-4)?

Mathematics
1 answer:
Airida [17]2 years ago
3 0

Answer:

B

Step-by-step explanation:

To solve -6-(-4) you need to break it down.

First, multiply -(-4) which would equal 4 (negative times negative equals positive)

Then add 4 to -6 (-6+4)

This equation is being expressed through option B

Hope this helps! :)

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HELP ASAP PLEAse!!! HOW DO YOU FIND THE LATERAL AREA OF THE TRIANGLE, (ALREADY GOT SURFACE AREA CORRECT), PLEASE ROUND IT TO THE
Zarrin [17]
172.9

You find it by multiplying each side of the triangle by the height (9) and adding the results together.

A(lateral) = 8 * 9.1 + 3 * 9.1 + 8 * 9.1
= 72.8 + 27.3 + 72.8
= 172.9
6 0
2 years ago
Help please!!!!!!????!!!!
zavuch27 [327]

Answer:

28 yd²

Step-by-step explanation:

find the two biggest sides

2x4

8 but there are two sides...

8*2= 16

*16*

now side panels

1x4

4 but there are two

4*2= 8

*8*

top and bottom panels

2x1

2 but there are two

2*2= 4

*4*

add them up

16 + 8 + 4= 28  

4 0
3 years ago
Solve for x in the equation;3x +34x -40x=0
saveliy_v [14]

Answer:

x=0

Step-by-step explanation:

3x + 34x - 40x = 0

37x - 40x = 0

-3x = 0

x=0

5 0
3 years ago
A) strongly negative
vlada-n [284]
B is the answer. PLEASE MARK AS BRAINLIEST!!!
7 0
3 years ago
Read 2 more answers
It is known that 60% of customers will need additional maintenance on their vehicle when coming in for an oil change. A random s
inn [45]

Answer:

the probability that more than 70% of customers in the sample will need additional maintenance is 0.0371

Step-by-step explanation:

From the information given:

we are to determine the probability that more than 70% of customers in the sample will need additional maintenance

In order to achieve that, let X be the random variable that follows a binomial distribution.

Then X \sim Bin(48, 0.6)

However 70% of 48 samples is

= 0.7 × 48 = 33.6 \simeq 34

Therefore, the required probability is:

= P(X> 34)

= \sum \limits ^{48}_{x=34} (^{48}_{x}) (0.6)^x(1 - 0.6)^{48-x}

= \dfrac{48!}{34!(48-34)!} (0.6)^{34} (0.4)^{48-34}

= 4.823206232 \times 10^{11} (0.60)^{34}(0.4)^{14}

= 0.03709524328

\simeq 0.0371

3 0
3 years ago
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