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2 years ago

7

Assume: Triangle CDE is a right triangle and Angle D is a right angle. What is the length of the missing side? A) 26 . B) 25 . C

) 24 . D) 23 .

1 answer:

kramer2 years ago

6 0

Answer:

Length of base DE = 24 units

Step-by-step explanation:

Given:

In given triangle, right angle at D

SO,

Perpendicular of given triangle = 32 unit

Hypotenuse of given triangle = 40 unit

Find:

Length of base DE

Computation:

Using Pythagoras theorem

Base = √Hypotenuse² - Perpendicular²

Length of base DE = √Hypotenuse of given triangle² - Perpendicular of given triangle²

Length of base DE = √40² - 32²

Length of base DE = √1,600 - 1,024

Length of base DE = √576

Length of base DE = 24 units

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In a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of th

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Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

Let Probability that the students passed the first exam = P(F) = 0.74

Probability that the students passed the second exam = P(S) = 0.72

Probability that the students passed both exams = $P(F \bigcap S)$ = 0.58

Now, if the student passed the first exam, probability that he passed the second exam is given by the conditional probability of P(S/F) ;

As we know that conditional probability, P(A/B) = $\frac{P(A\bigcap B)}{P(B) }$

Similarly, P(S/F) = $\frac{P(S\bigcap F)}{P(F) }$ = $\frac{P(F\bigcap S)}{P(F) }$ {As $P(F \bigcap S)$ is same as $P(S \bigcap F)$ }

= $\frac{0.58}{0.74}$ = 0.784

Therefore, probability that he passed the second exam is 0.784 .

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The ratio of the number of adults to the number of children in a library is 12:5. If there are 45 children in the library, how m

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3 years ago

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Step-by-step explanation:

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3 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

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Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

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

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

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 $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

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

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

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

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

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

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

So

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

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3 years ago