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Elis [28]
2 years ago
12

3. If a triangle has a height of 14 inches and a base of 9 inches, what's its area?

Mathematics
1 answer:
Strike441 [17]2 years ago
4 0
It’s b because 9x14/2= 63
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Question IS in PICTURE!!!!
julia-pushkina [17]
IQR= Q3-Q1

Airport 1
Q3 is 8
Q1 is 2

8-2= 6

Airport 1 IQR is 6

Airport 2
Q3 is 8
Q1 is 3

8-3= 5

Airport 2 IQR Is 5
8 0
3 years ago
Read 2 more answers
Find the area of the circle<br> Pls help quick
tigry1 [53]

Answer:

153.86.

Step-by-step explanation:

Formula for area of a circle: A= Pi R (R is radius) squared. First fill in the numbers: 3.14x7^2 Then Solve for the squared: 7^2=49, then finally multiply 49x3.14 which gets you to 153.86.

7 0
2 years ago
Oscar says to Vanesa: "I am five times the age you were when I was the age you are; and when you are the age I am, our ages will
Gemiola [76]

Answer:

Vanessa is 60 and the other person is 12

Step-by-step explanation:

X + 5x = 72

6x = 72

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3 0
2 years ago
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We have just moved to geometry. <br> Any help? <br><br> This is just practice as well
Vera_Pavlovna [14]

Answer:

rotation of 270 counterclock wise

Step-by-step explanation:

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8 0
2 years ago
There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit
Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a \frac{3}{11} chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is \frac{8}{11}, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is\frac{2}{11}, and the probability of picking the second red ball is \frac{1}{10} and the probability of picking the green ball is\frac{1}{9}. We want to multiply thm again, as per the multiplication rule like the last problem.

\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}

5 0
3 years ago
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