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2 years ago

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Prove that AABE = ADCE, given that AE = DE, BE CE, with the same vertex measure ZE.

1 answer:

mojhsa [17]2 years ago

8 0

Answer:

Prove that AABE = ADCE, given that AE = DE, BE CE, with the same vertex measure ZE.

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What is equivelant to 6 divided by 27

Sholpan [36]

The fractions that's equivalent to 6/27 is 2/9

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3 years ago

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Groups of adults are randomly selected and arranged in groups of three. The random variable x is the number in the group who say

Vera_Pavlovna [14]

Answer:

0.789

Step-by-step explanation:

First, find the expected value (the mean).

μ = ∑ x P(x)

μ = (0) (0.359) + (1) (0.436) + (2) (0.180) + (3) (0.025)

μ = 0.871

Next, find the variance.

σ² = ∑(x − μ)² P(x)

σ² = (0 − 0.871)² (0.359) + (1 − 0.871)² (0.436) + (2 − 0.871)² (0.180) + (3 − 0.871)² (0.025)

σ² = 0.622

Finally, find the standard deviation:

σ = √0.622

σ = 0.789

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3 years ago

Ten ounces of a liquid contain 20%

matrenka [14]

Answer:

A)20% because we added water not alcohol

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3 years ago

The initial mass of a certain species of fish is 6 million tons. The mass of fish, if left alone, would increase at a rate pro

Phoenix [80]

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Step-by-step explanation:

Check attachment for solution

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3 years ago

Darren invests $4,500 into an account that earns 5% annual interests. How much will be in the account after 10 years if the inte

Tom [10]

We have been given that Darren invests $4,500 into an account that earns 5% annual interests. We are asked to find the amount in his account after 10 years, if the interest rate is compounded annually, quarterly, monthly, or daily.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$5\%=\frac{5}{100}=0.05$

When compounded annually, $n=1$ :

$A=4500(1+\frac{0.05}{1})^{1\cdot 10}$

$A=4500(1.05)^{10}$

$A=4500(1.6288946267774414)$

$A=7330.025820498\approx 7330.03$

When compounded quarterly, $n=4$ :

$A=4500(1+\frac{0.05}{4})^{4\cdot 10}$

$A=4500(1.0125)^{40}$

$A=4500(1.6436194634870132)$

$A=7396.28758569\approx 7396.29$

When compounded monthly, $n=12$ :

$A=4500(1+\frac{0.05}{12})^{12\cdot 10}$

$A=4500(1.00416666)^{120}$

$A=4500(1.64700949769)$

$A=7411.542739605\approx 7411.54$

When compounded daily, $n=365$ :

$A=4500(1+\frac{0.05}{365})^{365\cdot 10}$

$A=4500(1.0001369863013699)^{3650}$

$A=4500(1.6486648137656943695)$

$A=7418.9916619456\approx 7419.00$

Since amount earned will be maximum, when interest is compounded daily, therefore, Darren should use compounded daily interest rate.

3 0

3 years ago