All categories

4 years ago

In need the proofs and statements

Mathematics

1 answer:

ozzi4 years ago

4 0

Answer:

Since we are given that angle Y = angle Z

The magnitude of the angles will also be equal

∠Y = ∠Z

(x + 10) = (2x - 90)

90 + 10 = 2x - x

100 = x

You might be interested in

Find the equation of the line, given the following information:

Art [367]

Answer:

Step-by-step explanation:

y - 4 = 5(x - 1)

y - 4 = 5x - 5

y = 5x - 1

3 0

4 years ago

A motor vehicle is____________.

brilliants [131]

A motor vechicle is motor cycle

7 0

2 years ago

Why cannot squares be placed in the hierarchy under trapezoids.

solong [7]

Because the are not the same size

3 0

3 years ago

Read 2 more answers

Multiply. Reduce to lowest terms.<br> 2 1/7 x 4 3/8

sveta [45]

2 1/7 x 4 3/8
15/7 x 3/2
45/14 or 3 3/14

7 0

3 years ago

Read 2 more answers

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

ss7ja [257]

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:

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

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

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

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.

3 0

3 years ago