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3 years ago

If EF = 5w + 20, FG = 7w + 28, and EG = 72, find the value of w. Find EF and FG

Mathematics

1 answer:

IrinaK [193]3 years ago

7 0

Answer:

w = 2, EF = 30 , FG = 42

Explanation:

We have EF = 5w + 20 and FG = 7w + 28

Since these points E, F and G are col-linear points we have.

EG = EF + FG

And given that EG = 72

So, EF+EG = 72

5w + 20 + 7w + 28 = 72

12w + 48 = 72

12 w = 24

w = 2

So EF = 5w + 20 = 5x2 + 20 = 30

FG = 7w + 28 = 7x2 + 28 = 42

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If a car traveled 162 miles in 3 hours, how many hours would it take to travel 351 miles? Please show work! :)

Flauer [41]

Answer:

390 minutes

Step-by-step explanation:

162 / 3 = 54

351 / 54 = 6.5

6.5 x 60 = 390 minutes

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3 years ago

Round each number to the nearest ten.

Bezzdna [24]

552.17 rounds to 550
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3 years ago

A family has two cats named Gordo and Flaco. Gordo weighs 15 pounds and Flaco weighs 8 pounds. A cat’s weight is classified as u

Elena-2011 [213]

Answer:

Gordo's weight = 15 pounds

It is outside the healthy weight range and is in the top 5% of weights of cats.

Gordo's weight makes Gordo unhealthy.

Step-by-step explanation:

μ = mean weight = 9.5 pounds

σ = standard deviation = 1.5 pounds

This is a normal distribution problem

We first calculate the limit of the bottom 5% of weights

Let the z-score for this limit be z'

P(z < z') = 0.05

From the normal distribution table,

z' = -1.645

And the limit for the top 5% which is z" = 1.645.

The weight that corresponds to these scores are then calculated.

Standardized scores are given as

z = (x - μ)/σ

So,

z' = (limit for the bottom 5% - μ)/σ

-1.645 = (limit for the bottom 5% - 9.5)/1.5

limit of the bottom 5% = (-1.645)(1.5) + 9.5 = 7.033 pounds

z" = ( (limit for the top 5% - μ)/σ

1.645 = (limit for the top 5% - 9.5)/1.5

limit of the bottom 5% = (1.645)(1.5) + 9.5 = 11.968 pounds

Therefore the healthy weight range for cats is (7.033 < x < 9.968)

Gordo's weight = 15 pounds

It is outside the healthy weight range and is in the top 5% of weights of cats.

Gordo's weight makes the cat unhealthy.

Flaco's weight = 8 pounds

Flaco's weight lies in the healthy weight range for cats. Hence, Flaco is a healthy cat.

8 0

3 years ago

Given the function, f(x)=10x-3, <br> what is the value of the function when x=-1/2

Mandarinka [93]

-8

f(-1/2)=10(-1/2)-3
f(-1/2)=(-5)-3
f(-1/2)=-8

6 0

3 years ago

A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters

natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used to solve this question.

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 combinations 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.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

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

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$

0.0479 = 4.79% probability that fewer than 11 of them will vote

8 0

3 years ago