All categories

4 years ago

Determine the growth factor corresponding to 1% increase

Mathematics

2 answers:

bezimeni [28]4 years ago

4 0

Not sure but I would say 0.1 again I am not sure that that is what is asking for

Juliette [100K]4 years ago

3 0

For an increase of percentage, it is a growth factor.

For a decrease of percentage, it is a decay factor.

The growth factor is = (1+r)

The decay factor is= (1-r)

As it is increase of 1%, it is a growth factor.

Here the rate of increase is 1% = 1/100 = 0.01

Hence the growth factor is 1 + 0.01 = 1.01

The growth factor corresponding to 1% increase is 1.01.

You might be interested in

7 times the sum of a number and 3

insens350 [35]

Answer:

Step-by-step explanation:

7x+3

3 0

4 years ago

What is the unit price for an 8 ounce package costing 1.59

IrinaVladis [17]

Image ?....?.?.?..?.?..?.?.?.?..?.?...?.?.?.?..??...?.?.??..?.?.? Smh simpsimpjdjs

4 0

3 years ago

Select the correct answer.

malfutka [58]

Answer:

Step-by-step explanation:

x - 4 squared provides 2 real zeros that are the same. These two are not distinct.

The other two come from x^2 - 7x + 10 which has a discriminate of

sqrt(7^2 - 4*1*10) = sqrt(9) = +/- 3 leading to something real and different.

6 0

3 years ago

Which equation represents the word sentence shown below?

Llana [10]

Answer:

where are the equations?

Step-by-step explanation:

tell me and I will edit my answer to answer ur question

7 0

3 years ago

Read 2 more answers

Using Newton's Law of Cooling. You are a medical examiner called to an abandoned house in which a dead body has been discovered.

Artyom0805 [142]

A) The differential equation comes from the fact that the rate of temperature change is proportional to the difference in temperatures.
$\frac{dT}{dt} = \alpha(T -52)$

B) Find general solution by separating variables and integrating $\int\frac{dT}{T-52} = \alpha \int dt \\ ln (T-52) = \alpha t + C \\ T = C e^{\alpha t} +52$ :

C) Initial condition is t=0, T = 87
$87 = C +52 \\ C = 35$

D) Total time elapsed is 10 minutes, new temperature is 84
$84 = 35 e^{10\alpha} +52$
solve for alpha
$e^{10\alpha} =\frac{32}{35} \\ \alpha = \frac{ln(32/35)}{10} \\ \alpha = -.00896$

E) Temperature function is:
$T = 35 e^{-.00896 t} +52$
solving for t
$t = \frac{ln (\frac{T-52}{35})}{-.00896}$
Plug in T = 98.6
$t = -31.95$

This is approximately 32 minutes before he arrived or about 1:20 AM

3 0

3 years ago