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3 years ago

Find the sum.

Mathematics

1 answer:

Strike441 [17]3 years ago

6 0

You are gonna use the distributive property.

4(2.5g - 4) + 3(1.2g - 2)

So you are going to multiply 4 with everything inside the first set of parenthesis and your gonna multiply 3 with everything insides the second set of parenthesis.

4 x 2.5g - 4 x 4 + 3 x 1.2g - 3 x 2

10g - 16 + 3.6g - 6

You wanna combine like terms:

10 - 3.6g + 16 - 6

6.4g + 10

You answer is 6.4g + 10.

Hope this Helps!!

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All three sides of a triangle are initially 5.0 long. One of the triangle's sides is oriented horizontally. The triangle is scal

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Answer:

h = 3.62

Step-by-step explanation:

First, as both triangles have the same angles we can use the relationship of areas and sides corresponding to similar triangles as follows:

$relation\ between\ areas=( relation\ between\ sides)^2\\\frac{0.7A}{A}=(\frac{s}{5})^2\\s= 5\sqrt0.7= 4.18$

Now we know that the new triangle has sides of 4.18. Then, as these triangles are equilateral we can use the Pythagorean Theorem to find the height:

$4.18^2=h^2+(\frac{4.18}{2})^2\\h=\sqrt (4.18^2-2.09^2)=3.62 m\\h= 3.62$

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2 years ago

Look at the number line below what point is marked on the number line

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Answer:

-2.5

Step-by-step explanation:

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2 years ago

2. CTfastrak bus waiting times are uniformly distributed from zero to 20 minutes. Find the probability that a randomly selected

Juliette [100K]

Answer:

b. 0.25

c. 0.05

d. 0.05

e. 0.25

Step-by-step explanation:

if the waiting time x follows a uniformly distribution from zero to 20, the probability that a passenger waits exactly x minutes P(x) can be calculated as:

$P(x)=\frac{1}{b-a}=\frac{1}{20-0} =0.05$

Where a and b are the limits of the distribution and x is a value between a and b. Additionally the probability that a passenger waits x minutes or less P(X<x) is equal to:

$P(X$

Then, the probability that a randomly selected passenger will wait:

b. Between 5 and 10 minutes.

$P(5$

c. Exactly 7.5922 minutes

$P(7.5922)=0.05$

d. Exactly 5 minutes

$P(5)=0.05$

e. Between 15 and 25 minutes, taking into account that 25 is bigger than 20, the probability that a passenger will wait between 15 and 25 minutes is equal to the probability that a passenger will wait between 15 and 20 minutes. So:

$P(15$