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mestny [16]
2 years ago
8

Pls help asap! i will mark brainliest!

Mathematics
1 answer:
jonny [76]2 years ago
4 0

Answer:

ok

Step-by-step explanation:

yes

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Help me plezzzzzzzzzzzzzzzzzz
alexdok [17]
31. There are C(5, 2) = 10 ways to choose 2 colors from a group of 10 colors if you don't care about the order. (Here, we treat blue background with violet letters as being indistinguishable from violet background with blue letters.)

Only one of those 10 pairs is "B and V", so the probability is 1/10.
   a) P(B and V) = 10%


32. The 12 inch dimension on the figure is 0 inches for the cross section. The remaining dimensions of the cross section are 
   c) 5 in. × 4 in.


_____
C(n, k) = n!/(k!·(n-k)!)
C(5, 2) = 5!/(2!·3!) = 5·4/(2·1) = 10
8 0
3 years ago
DUDE THIS IS MY LAST QUESTION PLEASE HELP
Alborosie

Answer:

B

Step-by-step explanation:

3 x -3 is -9

-2 x 4 is -8

-9 x -8 is 72

7 0
3 years ago
Read 2 more answers
Need help with this question.
Iteru [2.4K]
Take 3÷2.6=1.153846repeating. So around $1.15. To check you can take the decimal point times by 2.6 and it will give you three.
7 0
2 years ago
Help super important
const2013 [10]
Hello.

Taking a look at our screenshot provided, we can conclude that we need to find the missing angle degree out of 90 degrees, as we are dealing with a right angle.

Let's set this up as an Algebraic formula and solve for the variable;
5x + 15 + 50 = 90

First, let's combine like-terms (15 and 50).

5x + 65 = 90

Now, isolate our variable by subtracting 65 from each side of the equation.

90 - 65 = 25
65 - 65 = 0

5x = 25

Now, divide both sides by 5 to solve for x, our missing angle degree.

x = 5

Your answer is A.) 5

I hope this helps!
3 0
3 years ago
The large piston in a hydraulic lift has an area of 2m^2. What force must be applied to the small piston with an area of .2m^2 i
Helen [10]

Answer:

147,000N

Step-by-step explanation:

A1= 2m^2

A2= 0.2m^2

F2= 14,700N

Required

F1, the applied force

Applying the formula

F1/A1= F2/A2

substute

F1/2=14700/0.2

2*14700= F1*0.2

29400= F1*0.2

F1= 29400/0.2

F1=147,000N

Hence, the applied force is 147,000N

3 0
2 years ago
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