Choose the triangle that seems to be congruent to the given one.

the length of a rectngle is 8/3 times as great as its widthwhat is the width if the length of a rectangle is 24 ft.

laila [671]

L=(8/3)W = 24 ft. Solving for W, we mult. both sides of this eqn by (3/8), obtaining

W = (3/8)(24 ft) = 9 ft (answer)

4 0

3 years ago

Find the area of a circle who is diameter is 27 inches(use 3.1416)

Nimfa-mama [501]

Answer: The area is 572.5566

Step-by-step explanation:

27÷2=r=13.5

A=πr²

A=3.1416(π) · 13.5²(r)

A=572.5566

4 0

4 years ago

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Can 6 fit into 36 and if so how many times

ratelena [41]

Yes.
This is the same thing as:
$\frac{36}{6} =6$

How many times can 6 go into 36?
6 times
Try:
6*6=36

So true.

Hope this helps! :D

6 0

4 years ago

Read 2 more answers

For accounting purposes, the value of assets (land, buildings, equipment) in a business are depreciated at a set rate per year.

lidiya [134]

Answer: $ 125987.80

Step-by-step explanation:

Given: The value, V(t) of $393,000 worth of assets after t years, that depreciate at 15% per year, is given by the formula

$V (t)=V_o(b)^t$ , here $V_o$ is the initial asset value and b is the multiplicative decay factor.

The exponential decay function is given by ;-

$f(x)=A(b)^x$ , where A is the initial value , x is the times period and b is the multiplicative decay factor.

where b = 1-r, r is the rate of decay.

Since r = 15%=0.15

Therefore, b = 1-0.15=0.85

Now ,for 7 years , the value of assets is given by :-

$V=393000(0.85)^7=125986.795\approx125987.80$

Hence, the assets valued at after 7 years = $ 125987.80

4 0

3 years ago

Read 2 more answers

The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

<em>The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.</em>

5 0

3 years ago