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2 years ago

How do you simplify 2 15/10?

2 answers:

7 0

So you have to add another whole which is three then you take it away from the fifteen then you have 3 5/10 and simplify again and you get 4 1/2

sdas [7]2 years ago

6 0

First you take 15/10 and see that you can take 5 out of the top and bottom, so it would become 3/2. Now you have 2 3/2. 2 goes into 3 once with 1 remainder so 3/2= 1 1/2
So then you take the original 2 and add it to 1 1/2. The answer is now 3 1/2

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Angle M = 120 degrees

Shortest side = MN

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Step-by-step explanation:

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3 years ago

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2 years ago

Which expression represents twice the sum of w and 6?

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3 years ago

Angela asked each member of her class if they ever been to six flags over Georgia the class consists of 10 males and 14 females.

Feliz [49]

Answer:

2 male students didn't visit six falgs.

Step-by-step explanation:

Hello!

You have the following information:

The class has 10 male students and 14 female students.

From the females, 8 visited six flags over Georgia.

And from the total of students of class 16 visited the six flags over Georgia.

Using this information you have to calculate the number of male students that did not visit six flags.

First step: Calculate the number of male students that visited six flags.

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Second step: Calculate the number of students that didn't visit it.

From the total of male students in the class, 8 visited six flags so 10 - 8 = 2 male students didn't visit six flags over GA.

I hope this helps!

7 0

2 years ago

A box contains 5 tickets numbered 1,2,3,4, and 5. Two tickets are drawn at random from the box. Find the chance that the numbers

adelina 88 [10]

Answer:

(a) The probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) The probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

Step-by-step explanation:

The tickets are drawn such that the difference between the two numbers is at least 2.

That is, 1st number - 2nd number ≥ 2.

The sample space such that this condition is satisfied is:

S = {(1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5)} = 6 possible pairs.

But there is also case where the numbers can be drawn in reverse order, i.e we can draw (3, 1) instead of (1, 3).

This makes the total number of possible pairs as, 6 × 2 = 12 pairs.

(a) <u>With Replacement</u>

In case the tickets are selected with replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

$P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs$

$=\frac{1}{5}\times \frac{1}{5}\times12\\ =0.48$

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) <u>Without Replacement:</u>

In case the tickets are selected without replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

$P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs$

$=\frac{1}{5}\times \frac{1}{4}\times12\\ =0.60$

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

7 0

2 years ago