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3 years ago

Find the length of the radius. Assume that lines which appear to be tangent are tangent.

Mathematics

1 answer:

aleksley [76]3 years ago

3 0

The Answer Is 7.5

Tangent and 8 makes a right triangle, which is great because you could find the diameter and divide it by 2 to find the radius.

64+d^2=289

d^2=225

d=15

15/2=7.5

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State if each pair of ratios forms a proportion 12/24 and 3/4

raketka [301]

Answer:

Step-by-step explanation:

12/24 reduces to 1/2, whereas 3/4 does not reduce. This pair of ratios does not form a proportion.

If both ratios reduced to the same common value, that would represent a proportion.

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3 years ago

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A kangaroo hops 2 kiliometers in 3 minutes. At this rate: How far does the kangaroo travel in 6 minutes?

Bess [88]

Answer:

4km

Step-by-step explanation:

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3 years ago

8. A restaurant bill before tax is $15.60. How much is a 15% tip for this bill?

allsm [11]

Answer:

Step-by-step explanation:

the tip would be = 2.325 which rounds up to 2.33$ in total this would be17.83

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3 years ago

Mary brought a pair of jeans and a sweater. The pair of jeans costs $30 and the sweater $35. If sales tax is 6%, how much did Ma

Serjik [45]

Answer:

Total buy (without taxes) = 30 + 35 = 65 $

Taxes = 6%(35) = 2.1 $

Total Buy with Taxes = 65 +2.1 = 67.1 $

A: 67 dollars and 10 cents

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3 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

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

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

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

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

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

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

3 years ago