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2 years ago

Please help me! I will make you the Brainliest!

Mathematics

1 answer:

BARSIC [14]2 years ago

7 0

Answer: Is it C

Step-by-step explanation:

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Lines AB and CD (if present in the picture) are straight lines. Find x. Give reasons to justify your solutions.

MrRa [10]

Answer:

180- 110= 70

70+x+90=180

160+x=180

180-160=x

20=x

3 0

2 years ago

The angle of depression of top of a pole of

topjm [15]

Answer:

Didn't finish sorry. I wish I could help.

8 0

2 years ago

Find the midpoint between... 0 and 88

motikmotik

Answer:

i believe that the mid point is 44

8 0

3 years ago

Two of your friends go bowling. One friend rents a pair of bowling shoes for $3 and bowls 3 games. The other friend brings his o

Nesterboy [21]

Answer: The cost of each game is $2.50

Step-by-step explanation:

Let's find the equations for each friend:

Friend 1:

Pays $3 for the shoes, and plays 3 games, then if X is the cost of each game, Friend 1 pays a total of:

$3 + 3*X

Friend 2:

This friend buys a soda for $0.50, and he plays 4 games, remember that the cost of each game was X. then this friend pays:

$0.50 + 4*X

And we know that both friends pay the exact same amount, then we can write:

$3 + 3*X = $0.50 + 4*X

And solve this for X.

We need to isolate X, then we can move all the terms with X to the right, and all the terms without X to the left:

$3 - $0.50 = 4*X - 3*X

$2.50 = (4 - 3)*X = X

This means the cost of each game is $2.50

4 0

2 years ago

Suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the othe

Effectus [21]

Answer:

0.5798 or 57.98%

Step-by-step explanation:

The total number of ways to form the two teams is the combination of choosing 10 people out of 35 (₃₅C₁₀). The number of possibilities that A and B are both on the 10-people team is given by the combination of choosing 8 people (since two are fixed) out of 33 (₃₃C₈).The number of possibilities that A and B are both on the 25-people team is given by the combination of choosing 10 people out of 33 (₃₃C₈).

Therefore, the probability that two particular people A and B will be on the same team is:

$P = \frac{_{33}C_8+_{33}C_{10}}{_{35}C_{10}} \\\\P=\frac{\frac{33!}{(33-8)!8!}+\frac{33!}{(33-10)!10!} }{\frac{35!}{(35-10)!10!}} \\\\P=\frac{69}{119} = 0.5798$

The probability is 0.5798 or 57.98%.

8 0

3 years ago