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3 years ago

Please help with problem and show work

Mathematics

1 answer:

luda_lava [24]3 years ago

7 0

Answer:

x = 7

Step-by-step explanation:

The image below shows the relationship between secants chords and tangents.

Imagine that the letters in the image represent the numbers in the actually problem

E would equal 9

C would equal 12

and A would equal x

We would then use the same formula C² = E * ( E + A ) to solve for x , but once again instead of using letters we use the numbers in the problem

We would have 12² = 9( 9 + x )

we now solve for x

Step 1 simplify and distribute

12²=144

9 * 9 = 81

9 * x = 9x

we would then have 144 = 81 + 9x

step 2 subtract 81 from each side

144 - 81 = 63

81 - 81 cancels out

we now have 63 = 9x

step 3 divide each side by 9

63 / 9 = 7

9x / 9 = x

we're left with x = 7

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2 years ago

According to a 2018 survey by Bankrate, 20% of adults in the United States save nothing for retirement (CNBC website). Suppose t

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0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they save nothing for retirement, or they save something. The probability of an adult saving nothing for retirement is independent of any other adult. This means that the binomial probability distribution is used to solve this question.

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 combinations 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.

20% of adults in the United States save nothing for retirement (CNBC website).

This means that $p = 0.2$

Suppose that sixteen adults in the United States are selected randomly.

This means that $n = 16$

What is the probability that three or less of the selected adults have saved nothing for retirement?

This is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = C_{16,0}.(0.2)^{0}.(0.8)^{16} = 0.0281$

$P(X = 1) = C_{16,1}.(0.2)^{1}.(0.8)^{15} = 0.1126$

$P(X = 2) = C_{16,2}.(0.2)^{2}.(0.8)^{14} = 0.2111$

$P(X = 3) = C_{16,3}.(0.2)^{3}.(0.8)^{13} = 0.2463$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0281 + 0.1126 + 0.2111 + 0.2463 = 0.5981$

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

4 0

3 years ago