Explain the answer....

Even without doing the math, (and without having the requisite diagram to view), it seems likely that the rectangle was simply divided along the diagonal. And, if that's the case then the two triangles formed would have the same area. So, if one triangle's area ABC is 600 sq. yds. as the problem states, then the other triangle ACD also would be 600 sq. yds.

But if we do the math anyway, we can get the same result.

Keep in mind that:

P of rectangle = 2 ( l + w)

A of rectangle = (l x w)

A of triangle = h x b divided by 2

I'd say that the length of the rectangle is 40 and the width is 30.

P of rectangle is 2 (30 + 40) = 140 .... check

A of rectangle is 30 x 40 = 1,200 .... check

A of triangle is 30 x 40 divided by 2 = 600 ...... check

7 0

4 years ago

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A TV show, Lindsay and Tobias, recently had a share of 20, meaning that among the TV sets in use, 20% were tuned to that show. A

kati45 [8]

Answer:

8.6% probability that none of the households are tuned to Lindsay and Tobias.

There is a 91.4% probability that at least one household is tuned to Lindsay and Tobias.

There is a 32.2% probability that at most one household is tuned to Lindsay and Tobias.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are watching the TV show, or they are not. This means that we use the binomial probability distribution to solve this problem.

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:

$n = 11, p = 0.20$

Find the probability that none of the households are tuned to Lindsay and Tobias.

This is P(X = 0).

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

$P(X = 0) = C_{11,0}.(0.20)^{0}.(0.80)^{11} = 0.086$

8.6% probability that none of the households are tuned to Lindsay and Tobias.

Find the probability that at least one household is tuned to Lindsay and Tobias.

Either no household is tuned, or at least one is tuned. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.086 = 0.914$

There is a 91.4% probability that at least one household is tuned to Lindsay and Tobias.

Find the probability that at most one household is tuned to Lindsay and Tobias.

This is

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

$P(X = 1) = C_{11,1}.(0.20)^{1}.(0.80)^{10} = 0.236$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.086 + 0.236 = 0.322$

There is a 32.2% probability that at most one household is tuned to Lindsay and Tobias.

8 0

3 years ago

Explain the answer....​

Explain the answer....