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2 years ago

IF Triangle DEF JKL, find KL.

Mathematics

1 answer:

s344n2d4d5 [400]2 years ago

3 0

Answer:

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{KL}{EF}$ = $\frac{JK}{DE}$ , substitute values

$\frac{6x+3}{x+11}$ = $\frac{25}{10}$ ( cross- multiply )

10(6x + 3) = 25(x + 11) ← distribute parenthesis on both sides

60x + 30 = 25x + 275 ( subtract 25x from both sides )

35x + 30 = 275 ( subtract 30 from both sides )

35x = 245 ( divide both sides by 35 )

x = 7

Then

KL = 6x + 3 = 6(7) + 3 = 42 + 3 = 45 → b

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Diego was trying to write 2^3 · 2^2 with a single exponent and wrote 2^3·2^2 = 2^3*2 = 2^6 . Explain to Diego what his mistake w

cupoosta [38]

Answer:

2^3·2^2 = 2^3+2 = 2^5

Step-by-step explanation:

Diego was trying to write 2^3 · 2^2

He wrote 2^3·2^2 = 2^3*2 = 2^6

But this is wrong because when bases are same exponents are added.

This is the law of exponents.

The correct form would be

2^3·2^2 = 2^3+2 = 2^5

For understanding it better we can write it like this

2^3·2^2 =

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3 years ago

If side b is 12 and side c is 31. Round your answer to the nearest hundredth.

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Answer:

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Step-by-step explanation:

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3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial 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:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

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

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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3 years ago

Which statement must be true about line TU?

just olya [345]

Line tu is parallel to rs

5 0

3 years ago

Read 2 more answers

A survey of 100 high school students provided this

HACTEHA [7]

Answer:

35/100

Step-by-step explanation:

You have to find how many juniors there are, so if you add 13, 20, and 2 you get 35. For the denominator you have to find the total number of students, so just add all of the numbers together to get 100. So there is a 35/100 chance that a randomly selected student is a junior.

5 0

1 year ago