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2 years ago

Which of these was not a factor in the build up to World War I?

Mathematics

2 answers:

const2013 [10]2 years ago

7 0

Answer:

A.) <u>A familial dispute amongst monarchs as to who inherited territory in the Balkans</u>

Explanation:

I took the test.

sammy [17]2 years ago

4 0

Answer:

A familial dispute amongst monarchs as to who inherited territory in the balkans.

Step-by-step explanation:

I learned from my mistake and it gave me that this was the answer.

Goodluck

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I WILL GIVE BRAINLIEST!!! 10 POINTS!!!

Natasha_Volkova [10]

Answer:1

Step-by-step explanation: 1/2 + 1/2 is 1

8 0

2 years ago

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Question is attached, ill mark brainliest

Alborosie

Answer:

w=7

Step-by-step explanation:

3w+4w=49

7w=49

w=49/7

w=7

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2 years ago

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What is 9/10-3/5 in fractions

Delvig [45]

-3/5 = -6/10
9/10 - 6/10 = 3/10

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2 years ago

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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

Find the midpoint of the segment with the following end points (10,5) and (6,9)

Taya2010 [7]

Answer:

The mid-point between the endpoints (10,5) and (6,9) is:

$\left(x,\:y\right)=\left(8,\:7\right)$

Step-by-step explanation:

Let (x, y) be the mid-point

Given the points

(10,5)

(6,9)

Using the formula to find the mid-point between the endpoints (10,5) and (6,9)

$\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$

Here:

$\left(x_1,\:y_1\right)=\left(10,\:5\right),\:\left(x_2,\:y_2\right)=\left(6,\:9\right)$

Thus,

$\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$

$\left(x,\:y\right)=\left(\frac{6+10}{2},\:\frac{9+5}{2}\right)$

$\left(x,\:y\right)=\left(8,\:7\right)$

Therefore, the mid-point between the endpoints (10,5) and (6,9) is:

$\left(x,\:y\right)=\left(8,\:7\right)$

3 0

2 years ago