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If triangle CDE is dilated by a scale factor of 1/5 with a center of dilation at vertex E, what is the area of triangle C'D'E'?

timurjin [86]

The correct answer would be Choice B: 4 square units.

When the scale factor is 1/5, that means the lengths of the sides are 1/5 of the original size. So instead of having a base of 20 and height of 10, the new triangle has a base of 4 and a height of 2.

The area of the triangle is: (4 x 2) / 2 = 4

8 0

3 years ago

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All points on a given line have their y-coordinate as 9. What is the equation of this line? A. x + y = 9 B. x – y = 9 C. x = 9 D

Phoenix [80]

D. y = 9 because it has a complete zero slope while the y-intercept of the equation is 9, so the equation is y = 9.

4 0

2 years ago

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Geometry grade 10 how do we see the adjecent side

RideAnS [48]

Answer: These sides are labeled in relation to an angle. The opposite side is across from a given angle. The adjacent side is the non-hypotenuse side that is next to a given angle.

Step-by-step explanation:

8 0

1 year ago

I need help with this problem

Serga [27]

It’s 0.09 or .09. Hope this helps:)

3 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

2 years ago