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Natalija [7]
3 years ago
12

15 POINTS! THANKS ON QUESTION AND PROFILE! 5 STAR RATING! FIRST ANSWER GETS THE BRAINLIEST!

Mathematics
2 answers:
Mamont248 [21]3 years ago
5 0
The correct answer is the bottom one
DIA [1.3K]3 years ago
3 0
The right answer is option D

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Find the amount of sap the maple syrup producer would have collected from each tree if the total amount (5 gallons) was redistri
Svetllana [295]

It's 1/2 gallon. Why? well, if you're doing I-Ready I suppose you seen 10 X's on that line plot. 10 divided by 5= 2 which is also one half. Why is it one half? Because half of ten is 5, and since I said I thought you were doing I-Ready, you would have to put it in a fraction.

5 0
3 years ago
2. Travis buys 2 quarts of milk at the store. What would be the equivalent or same as in
Vinvika [58]

Answer:

4 pints

Step-by-step explanation:

There are 2 pints in a quart, and there are 2 quarts, so 2 x 2 is 4.

3 0
3 years ago
Lulu bought 43 special lemons for making
scoray [572]

Answer:

x+43=29.24

Step-by-step explanation:

4 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
IF Tyler s currently 5 inches
FinnZ [79.3K]

Answer:

7 more inches to be a foot tall

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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