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tankabanditka [31]
3 years ago
6

What's 2|–9| – |–2|? Question 14 options: A) –18 B) –20 C) 16 D) 20

Mathematics
1 answer:
maksim [4K]3 years ago
3 0

16

find the absolute values ie. 9 and 2

multiply 9 by 2 and subtract 2 by 18

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Imani spent half of her weekly allowance playing mini-golf. To earn more money her parents let her wash the car for $4. What is
Pepsi [2]

Answer:

Well all you have to do is 4x=12 but muiltuply whatever numbers add to 12.such as 3x4=12

Step-by-step explanation:

5 0
3 years ago
(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

5 0
3 years ago
How to find the solution for 3x-11=x+3 ?​
Marina CMI [18]

Answer:

7

3 times ? - 11

=

? + 3

? = 7

Step-by-step explanation:

3 times 7 is 21 -11 is 10

7+3 is 10

Both equations has to make 10 so it is the answer of 7

Hope this helps :)

3 0
3 years ago
Read 2 more answers
Find the missing length.<br> 8<br> 3<br> с<br> C =<br> = ✓ [?]<br> Pythagorean Theorem: a2 + b2 = 2
Keith_Richards [23]

Answer:

✓73 or approximately 8.54

Step-by-step explanation:

✓ 8²+3² = ✓ 73 or approximately 8.54

4 0
3 years ago
What’s the answer to the math problem? (3x+2)(2x2+x+3
sergey [27]

{6x}^{3}  +  {7x}^{2}  + 11x + 6
factor out the function
8 0
4 years ago
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