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qaws [65]
3 years ago
7

List the laws of solid friction​

Mathematics
2 answers:
Artist 52 [7]3 years ago
7 0

Answer:

I WANT MY POINTS BACK

Step-by-step explanation:

FinnZ [79.3K]3 years ago
5 0
* The force of friction acts in opposite direction in which the surface is having tendency to move .
* the force of friction is equal to force applied to surfaces , so long as surface is at rest .
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Step-by-step explanation:

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1 4/9 divided by -2/9
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Answer:

6 1/2

Step-by-step explanation:

change 1 4/9 into a improper fraction, which is 13/9, and when dividing fractions flip the second number upside down then multiply. 13/9 times -2/9 is -117/18. simplify, which is 13/2, which is also equal to 6 1/2. hope this helped

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3 years ago
Select the student who reads the fastest (greatest number of words per minute)
kirill115 [55]
Im pretty sure it would be Ariel.
Tamar reads 600 words in 5 minutes, so he is already out of the question since Levi reads 700 words in 5 minutes. (Levi reads 140 words in 1 minute, so add 140+560=700 because they read 560 in 4 minutes)
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3 years ago
Jed made 2 free throws in 5 second how many would he make in one minute
Nataly_w [17]
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3 years ago
Read 2 more answers
A stack of cards consists of seven red and four blue cards. A second stack of cards consists of eleven red cards. A stack is sel
Ann [662]

Answer:

0.1733 = 17.33% probability the first stack was selected.

Step-by-step explanation:

To solve this question, it is needed to understand conditional probability, and the hypergeometric distribution.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Conditional probability:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

Probability of all red cards for the first stack:

For this, we use the hypergeometric distribution, as the cards all chosen without replacement.

7 + 4 = 11 cards, which means that N = 11.

7 red, which means that k = 7

3 are chosen, which means that n = 3

We want all red, so we find P(X = 3).

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 3) = h(3,11,3,7) = \frac{C_{7,3}*C_{4,0}}{C_{11,3}} = 0.2121

Conditional probability:

Event A: All red

Event B: From the first stack.

Probability of all red cards:

0.2121 of 50%(first stack)

1 of 50%(second stack). So

P(A) = 0.2121*0.5 + 1*0.5 = 0.60605

Probability of all red cards and from the first stack:

0.21 of 0.5. So

P(A \cap B) = 0.21*0.5 = 0.105

What is the probability the first stack was selected?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.105}{0.60605} = 0.1733

0.1733 = 17.33% probability the first stack was selected.

4 0
3 years ago
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