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3 years ago

I need help with #1 and #2 how do I do this?

Mathematics

2 answers:

Natalija [7]3 years ago

5 0

Answer:

The answer to your question is below

Step-by-step explanation:

Data

1) AB = 6 BC = 4 AE = 9 ED = ?

Use proportions to solve this problem

AB / AE = 6/9 (AB + BC) / (AE + ED) = (6 + 4) / (9 + ED)

-Equal both proportions

6/9 = 10 / (9 + ED)

-Solve for ED

9 + ED = 10(9)/ 6

-Simplification

9 + ED = 90/6

9 + ED = 15

ED = 15 - 9

-Result

ED = 6

2.- AB = 12 AC = 16 ED = 5 AE = ?

Use proportions to solve this problem

AE / AB = AE / 12 AD/AC = (AE + 5) / 16

-Equal both proportions

AE/12 = (AE + 5 ) / 16

-Solve for AE

16AE = 12(AE + 5)

16AE = 12AE + 60

16AE -12AE = 60

4AE = 60

AE = 60/4

-Result

AE = 15

Dafna11 [192]3 years ago

3 0

Answer:

Step-by-step explanation:

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2 years ago

Basil earned 631.40 in 7 years on an investment at a 5.5% simple interest rate. How much was basils investment

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3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 – 5t2, where t is the tim

Nostrana [21]

Answer:

$\sqrt{2}$ - 1

Step-by-step explanation:

h = 20 – 5t2 = 15 -> t = 1

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2 years ago

Read 2 more answers

6/12<br> is that equivalent to<br> 1/12

prohojiy [21]

No 6/12 is equal to 12/24

5 0

3 years ago

(a) A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure

Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

=1−e−1.21=0.703

The resultant probability is 0.703

Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

8 0

2 years ago