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Umnica [9.8K]
2 years ago
15

In this diagram, ABAC – AEDF. If the area of ABAC = 6 in?, what is the area of AEDF?

Mathematics
1 answer:
shepuryov [24]2 years ago
3 0

Answer:

2.7 in²

Step-by-step explanation:

similar triangles have the same angles, and all side lengths (or other distances) of one triangle have the same scaling factor to the side lengths of the other triangle.

so, we know the relation between the 2 baselines is 2/3, as this is the factor to turn the baseline of the large triangle into the baseline of the smaller triangle.

in other words

EF = BC × 2/3

2 = 3 × 2/3

correct

how do we calculate the area of a triangle ?

Area = baseline × height / 2

from BAC we know

Area = 6

baseline = 3

height = ?

6 = 3 × height / 2

12 = 3 × height

height = 4

aha !

now, EDF has a height too that we need to calculate is Area. and this height has the same scaling factor compared to the larger triangle as the side lengths : 2/3

so, for EDF we know

Area = ?

baseline = 2

height = 4 × 2/3 = 8/3

therefore, the area is

Area = (2 × 8/3) / 2 = (16/3) / 2 = 8/3 = 2.66666... ≈ 2.7

the shirt answer would be :

we know from the 2 baselines that the scaling factor for each distance is 2/3.

for the area we need to multiply 2 distances, so that means we have to multiply both by 2/3. and so on the formula for the area we have to use 2/3 × 2/3.

2/3 × 2/3 = 4/9

=>

Area small = Area large × 4/9 = 6 × 4/9 = 24/9 = 8/3 ≈ 2.7

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Answer:

d) The highest probability occurs when x equals 0.7500

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability, given by the following formula:

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.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

In this problem, we have that:

n = 15, p = \frac{1}{20}

a) The standard deviation is 0.8441

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15*0.05*0.95} = 0.8441

This is correct

b) The number of trials is equal to 15

n is the number of trials and n = 15. So this option is correct.

c) The probability that x equals 1 is 0.3658

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

P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658

This option is correct.

d) The highest probability occurs when x equals 0.7500

False. The number of sucesses is a discrete number, that is, 0, 1, 2,...,15. P(X = 0.75), for example, does not exist.

e) The mean equals 0.7500

E(X) = np = 15*0.05 = 0.75

This option is correct.

f) None of the above

d is false

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