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2 years ago

Given: EL- tangent, EK- secant Prove: EJ·LK = EL·LJ

Mathematics

1 answer:

drek231 [11]2 years ago

3 0

Answer:

This is possible.

Step-by-step explanation:

We can say that m<E=m<E, because of the Reflexive Property

Then, we have angles JKL and ELJ, which are equal through the peripheral angle theorem.

With these two angles, we can say that triangles ELK and EJL are similar, by the Angle-Angle Postulate (AA).

Then we can create this ratio through the Corresponding Parts of Similar Triangles Theorem, (CPST), $\frac{LK}{LJ} =\frac{EL}{EJ}$ .

With this ratio, we can cross multiply to get the desired result

$EJ$ · $LK=EL$ · $LJ$

Hope this helps with your RSM problem

Yup, i caught ya.

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A rectangular pool 18 meters by 12 meters is surrounded by a walkway of width x

vesna_86 [32]

Answer:

x=3 meters

Step-by-step explanation:

step 1

Find the area of the rectangular pool

$A=LW$

we have

$L=18\ m\\W=12\ m$

substitute

$A=18(12)=216\ m^2$

step 2

Find the area of rectangular pool including the area of the walkway

Let

x ----> the width of the walkway

we have

$L=(18+2x)\ m\\W=(12+2x)\ m$

substitute

$A=(18+2x)(12+2x)$

step 3

Find the area of the walkway

To find out the area of the walkway subtract the area of the pool from the area of rectangular pool including the area of the walkway

$A=(18+2x)(12+2x)-216$

step 4

Find the value of x if the area of the walkway equal the area of the pool

$(18+2x)(12+2x)-216=216$

Solve for x

$(18+2x)(12+2x)=432\\216+36x+24x+4x^{2}=432\\4x^{2} +60x-216=0$

Solve the quadratic equation by graphing

The solution is x=3 meters

see the attached figure

8 0

2 years ago

Leon bought trail mix that cost $0.78 per pound. He paid $6.24 for the trail mix. How many pounds of trail mix did he buy?

12345 [234]

If Leon bought trail mix worth of $ 6.24, and he paid 0.78 per pound,

6.24/0.78= 8 lbs

3 0

1 year ago

Read 2 more answers

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

ASHA 777 [7]

Answer:

a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of 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:

Google Chrome has a 20.37% share of the browser market. This means that $p = 0.2037$

20 Internet users are sampled, so $n = 20$ .

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

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

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050$

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

$E(X) = np$

$E(X) = 20*0.2037 = 4.074$

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

$Var(X) = np(1-p)$

$Var(X) = 20*0.2037*(0.7963) = 3.2441$

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

$\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011$

The standard deviation for the number of Chrome users is 1.8011.

6 0

2 years ago

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so $n = 1210$ .