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Vlad [161]
3 years ago
8

Is this a right triangle?Use the Pythagorean Theorem to find out! [picture above]​

Mathematics
1 answer:
S_A_V [24]3 years ago
8 0
No, 10^2 + 14^2 ≠ 18^2
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Pls answer and get branily
Dennis_Churaev [7]

Answer:

Just put 4 basketballs and 5 baseballs in each one

Step-by-step explanation:

If you think about it you will know that each ratio there is the same. So, you do the same thing each time.

All you have to do now is read. It says put 4 basketballs and 5 baseballs.

I hope this helped ;)

6 0
3 years ago
A scale measures weight to the nearest 0.1 pounds. Which is the most appropriate way to report weight using this scale?
Pani-rosa [81]
The question is kind of vague in my opinion.

the best measurement to report with the scale in my opinion would be tens of pounds. my reasoning is because the decimal place would still prove importance with numbers under one hundred. above one hundred pounds and the decimal place becomes kind of pointless. 

I hope I answered your question. though i'm sorry if i didn't 
8 0
3 years ago
List all the multiples of 4 up to 20
emmainna [20.7K]
4 , 8 , 12 , 16 , 20
3 0
3 years ago
Find the LCM of 3,12,16​
maria [59]

Answer:

The lcm is 48. Hope this helps

7 0
3 years ago
Read 2 more answers
There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl
insens350 [35]

We're told that

P(A\cap B)=0.15

P(A\cup B)^C=0.06\implies P(A\cup B)=0.94

P(B\mid A)=P(B^C\mid A)=0.5

where the last fact is due to the law of total probability:

P(A)=P(A\cap B)+P(A\cap B^C)

\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)

\implies 1=P(B\mid A)+P(B^C\mid A)

so that B\mid A and B^C\mid A are complementary.

By definition of conditional probability, we have

P(B\mid A)=P(B^C\mid A)

\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}

\implies P(A\cap B)=P(A\cap B^C)

We make use of the addition rule and complementary probabilities to rewrite this as

P(A\cap B)=P(A\cap B^C)

\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)

\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)

\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]

\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]

\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C

\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)

By the law of total probability,

P(B)=P(A\cap B)+P(A^C\cap B)

\implies P(A^C\cap B)=P(B)-P(A\cap B)

and substituting this into (*) gives

2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]

\implies P(B)=P(A\cup B)-P(A\cap B)

\implies P(B)=0.94-0.15=\boxed{0.79}

8 0
3 years ago
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