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2 years ago

Darcy keeps $5,000 in a bank account that does

Mathematics

1 answer:

den301095 [7]2 years ago

4 0

Answer:

Amount withdrawal = $2,847.66 (Approx)

Step-by-step explanation:

Given:

Amount deposit = $5,000

Withdraws percent = 10 % = 0.1

Number of weeks = 8

Find:

Amount withdrawal

Computation:

Amount withdrawal = 5,000[1-(0.1)⁸]

Amount withdrawal = 5,000[0.56953279]

Amount withdrawal = $2,847.66 (Approx)

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Find the exact values for the lengths of the labeled segments a, b and p drawn in green, red, and blue, respectively. Note that

adell [148]

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2 years ago

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An engineering student answered 81 questons correctly on a 90 question test. What percent of the question did she answer correct

anyanavicka [17]

Answer:

90% of the questions she answered correctly

Step-by-step explanation:

%age = 81/90 * 100= 90 % correctness

6 0

2 years ago

For a certain type of copper wire, it is knownthat, on the average, 1.5 flaws occur per millimeter.Assuming that the number of f

mario62 [17]

Answer:

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters

Step-by-step explanation:

Step 1:-

Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.

by Poisson random variable given that λ = 1.5 flaws/millimeter

Poisson distribution $P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}$

Step 2:-

The probability that no flaws occur in a certain portion of wire

$P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}$

On simplification we get

P(x=0) = 0.223 flaws occur / millimeters

Conclusion:-

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters

5 0

2 years ago

Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont

castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid is added to jar A.

Now jar A contains 4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution = [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50% as it is given that jar A has 50% at the end.

7 0

2 years ago

A plane is flying at 1000 meters above the ground and fires a laser 60 degrees of of straight down. what is the distance the lig

juin [17]

Answer:

Step-by-step explanation:

The distance the light traveled to the ground is the range of the plane expressed as:

R = U*√2H/g

U is he velocity

H is the max height

g is the acc. due to gravity

Let us get U first using the formula

H = u²sin²Ф/2g

1000 = u²sin²60/2(9.81)

1000*19.62 = 0.8660u²

19620 = 0.8660u²

u² = 19620/0.8660

u² = 22655.88

u = √22655.88

u = 150.52m/s

Next is to get the range

R = 150.52√2(1000)/9.81

R = 150.52 √2000/9.81

R = 150.52√203.87

R = 150.52*14.28

R = 2149.16m

Hence the distance the light traveled to the ground is 2,149.16m

5 0

2 years ago