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2 years ago

PLS HELP! Whoever gets it right will get brainliest!!

Mathematics

2 answers:

Anna11 [10]2 years ago

8 0

X= (-5/2) or X= 2.5 or X=2 1/2

melomori [17]2 years ago

7 0

Answer:

Step-by-step explanation:

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120 times 1.75 show your work plz help me I really dont get it

sineoko [7]

120 * 1.75 = 200

5 times 0 = 0

5 times 2 = 10 carry the 1

5 times 1 = 5 + 1 = 6

7 times 0 = 0

7 times 2 = 14 carry the 1...

7 times 1 = 7 + 1 = 8
1 times 0 = 0
1 times 2 = 2
1 times 1 = 1
add it all together and you get 200.

6 0

3 years ago

What is the range of possible sizes for side x?<br> 3.2<br> x<br> 5.5

ladessa [460]

Answer: try multiplication for the answer or use photomath

Step-by-step explanation:

3 0

2 years ago

Help please asap ??!!?

vampirchik [111]

Answer:

Religious Officials

This could be wrong, but I hope it helps!

4 0

2 years ago

Anyone know how to do #21?

Andreas93 [3]

Would it be y2 ? idk im not sure but i tried

6 0

3 years ago

20 people applied for a job. Everyone either has a school certificate or diploma or even both. If 14 have school certificates an

STatiana [176]

Given:

Either has a school certificate or diploma or even both = 20 people

Having school certificates = 14

Having diplomas = 11

To find:

The number of people who have a school certificate only.

Solution:

Let A be the set of people who have school certificates and B be the set of people who have diplomas.

According to the given information, we have

$n(A)=14$

$n(B)=11$

$n(A\cup B)=20$

We know that,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=14+11-n(A\cap B)$

$20=25-n(A\cap B)$

Subtract both sides by 25.

$20-25=-n(A\cap B)$

$-5=-n(A\cap B)$

$5=n(A\cap B)$

We need to find the number of people who have a school certificate only, i.e. $n(A\cap B')$ .

$n(A\cap B')=n(A)-n(A\cap B)$

$n(A\cap B')=14-5$

$n(A\cap B')=9$

Therefore, 9 people have a school certificate only.

3 0

2 years ago