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2 years ago

3

Bob bought a bike for $250 and repaired it. He can now sell the bike for $300. What is the percent increase in the value of Bob'

s bike?

1 answer:

rewona [7]2 years ago

4 0

Answer:

20% increase.

Step-by-step explanation:

300/250 = 1.2

You might be interested in

You have just received an inheritance of $28,000 and would like to invest it into an account. The bank offers two investment pla

grigory [225]

First we need to calculate annual withdrawal of each investment
The formula of the present value of an annuity ordinary is
Pv=pmt [(1-(1+r)^(-n))÷(r)]
Pv present value 28000
PMT annual withdrawal. ?
R interest rate
N time in years
Solve the formula for PMT
PMT=pv÷[(1-(1+r)^(-n))÷(r)]

Now solve for the first investment
PMT=28,000÷((1−(1+0.058)^(−4))
÷(0.058))=8,043.59
The return of this investment is
8,043.59×4years=32,174.36

Solve for the second investment
PMT=28,000÷((1−(1+0.07083)^(
−3))÷(0.07083))=10,685.63
The return of this investment is
10,685.63×3years=32,056.89

So from the return of the first investment and the second investment as you can see the first offer is the yield the highest return with the amount of 32,174.36

Answer d

Hope it helps!

6 0

2 years ago

Find the exact value of cos A in simplest radical form?

Mamont248 [21]

Answer:

cosA = 7/8

Step-by-step explanation:

cos∅ = adjacent over hypotenuse

cosA = 14/16

cosA = 7/8

3 0

2 years ago

The radius of a circular pond is increasing at a constant rate, which can be modeled by the function , where t is time in months

horrorfan [7]

Answer:

B. A(r(t)) = 25πt²

Step-by-step explanation:

Find the completed question below

The radius of a circular pond is increasing at a constant rate, which can be modeled by the function r(t) = 5t where t is time in months. The area of the pond is modeled by the function A(r) = πr². The area of the pond with respect to time can be modeled by the composition . Which function represents the area with respect to time? A. B. C. D.

Given

A(t) = πr²

r(t) = 5t

We are to evaluate the composite expression A(r(t))

A(r(t)) = A(5t)

To get A(5t), we will replace r in A(t) with 5t and simplify as shown

A(5t) = π(5t)²

A(5t) = π(25t²)

A(5t) = 25πt²

A(r(t)) = 25πt²

Hence the composite expression A(r(t)) is 25πt²

Option B is correct.

5 0

2 years ago

A geyser is a hot spring that periodically erupts a mixture of hot water and steam. Old Faithful in Yellowstone National Park is

vfiekz [6]

Answer:

Check the explanation

Step-by-step explanation:

Open file in excel.

Click on Data->Data Analysis and select Regression.

Enter wait time data range in y-variable.

Enter Duration data range in x-variable.

Click ok.

From above output:

#1 The equation of the linear regression line is:

A 95% confidence interval for the slope of the regression line is:

${(10.110,11.349)}$

In excel, ise formula =FORECAST(6,wait time data range, duration data range) and press enter. The wait time between eruptions following a 6 minute eruption is $\hat{Y}_{x=6}= 97.9} .$

Since (for F test) $p-value=0.000$ , we can conclude that there is a statistically significant linear relationship between the duration of the eruptions and the wait time between eruptions.

The correct null and alternative hypotheses for this analysis:

${ H_0:\beta_1=0 \ H_a:\beta_1\ne 0}$

Kindly check the output below

4 0

2 years ago

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar

Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 0

2 years ago