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Artyom0805 [142]
3 years ago
6

Some advantages of consolidating your debt into a single loan are _________. Help Please !

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
8 0
Well, one advantage simply is that its a loan.  If its a loan, you can pay little by little until its paid off.  Not sure if this helps, but I hope it does.  
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3 because 21/3 is 7. 24/3 is 8. And 27/3 is 9.
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Crazy T’s T-shirts can make 3/12 of a sweatshirt order in 1/6 of an hour. How much of an order can Crazy T complete in 1 hour?
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1.5 sweaters or 1 and half of a sweater

Step-by-step explanation:

you would multiply 6 by 3/12 and get 1 and a half

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A right triangle has side length of 4 cm and 5 cm what is the length of hypotenuse?
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\sqrt{41}

Step-by-step explanation:

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3 years ago
Sophia has 7m of ribbon to help decorate a room. She will cut pieces of a ribbon that are each r inches long. (1in = 2.54cm)
m_a_m_a [10]
I’m pretty sure it’s, D, if not, it’s C, I haven’t done this in a few years so correct me if I’m wrong, but I’m more sure it’s D.
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2 years ago
"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single d
dem82 [27]

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

In this problem we have that:

There are 6 possible outcomes for the dice. This means that the probability that it is a '3' is \frac{1}{6} = 0.167

There are 10 throws, so n = 10.

Probability of throwing AT LEAST two ‘3’s on the dice?

Either you throw less than two, or you throw at least two. The sum of the probabilities of these events is 1. So

P(X < 2) + P(X \geq 2) = 1

P(X \geq 2) = 1 - P(X < 2).

In which

P(X < 2) = P(X = 0) + P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.167)^{0}.(0.833)^{10} = 0.1609

P(X = 1) = C_{10,1}.(0.167)^{1}.(0.833)^{9} = 0.3225

So

P(X < 2) = P(X = 0) + P(X = 1) = 0.1609 + 0.3225 = 0.4834.

Finally:

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4839 = 0.5161.

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

5 0
2 years ago
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