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Degger [83]
3 years ago
11

The height of a triangle is 2cm more than the base of the height is increased by 2cm while the base remains the same the new are

a becomes 85.5cm^2 find the base and height of original triangle
Mathematics
1 answer:
san4es73 [151]3 years ago
4 0

Answer:

Remember that a triangle's area is bh/2, where

b = base

h = height

So. You start with a triangle of base b and height h, add 2 to the height, and get a triangle of area 82.5.

b(h+2)/2 = 82.5

b(h+2) = 165

What you want to do from here is test the b and h from the choices given, and figure out which b and h satisfy this.

(a) If b=13 and h=15, b(h+2) = 221 nope.

(b) If b=9 and h=11, b(h+2) = 117 nope.

(c) If b=11 and h=13, b(h+2) = 165 that's it!

Step-by-step explanation:

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MissTica

Answer:

(-5 , 0) and (6 , 0)

Step-by-step explanation:

In order to get the coordinates of the x-intercept(s) of the graph of y=(x-6)(x+5)

we need to solve for x the equation (x-6)(x+5) = 0

(x-6)(x+5) = 0

⇌

x - 6 = 0  or  x + 5 = 0

⇌

x = 6  or  x = -5

therefore

the coordinates are :

(-5 , 0) and (6 , 0)

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3 years ago
I Need Help Only #8
Sergeeva-Olga [200]

Answer:

15

Step-by-step explanation:

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The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,
professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so \mu = 7*0.6 = 4.2

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

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

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150

P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630

P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

3 0
3 years ago
PLZ HELP NEED ANSWER ASAP! A sports club rewards teams based on overall points earned in a season. The data for points are shown
uysha [10]
<span>You are given the following data and in order to answer the questions below, base your analysis in the given data.

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Part A: If the club wants to award the team that has the most consistent scoring among its team members, the team that should it choose is Team B because it has only lesser points if it scored low, at 38 and it has the next to the highest score among the three team, which is 49. Also, the range of their performance is 43.9, a much better score than 42.1 in Team A and 31.8 in Team C. 

Part B: If the club wants to award the team with the highest average score, </span>Team A because it has only lesser points if it scored low, at 22 and it has the highest score among the three team, which is 58. 
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What is 4.583.269 rounded to the nearest hundredth
Ede4ka [16]

4,583.27 is 4,583.269 rounded to the nearest hundredth



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