Ask question

Ask question

All categories

3 years ago

7

The price of 1 gallon of milk has increased $3.24 to $4.05. What is the percentage

1 answer:

AURORKA [14]3 years ago

5 0

4.05 - 3.24 = 0.81
(0.81 / 3.24)*100 = 25%

You might be interested in

The mean of birth weights for individual babies in the population is 3,500 grams. What is the mean birth weight for the sampling

Gala2k [10]

Answer:

The mean birth weight for the sampling distribution is

3,500 grams.

Step-by-step explanation:

The sample mean is the average of the sample values collected divided by the number of the samples, while the population mean is the average or mean of all the values in the population. If the sample is random and the sample size is large enough, then the sample mean would be a good estimator of the population mean. This implies that with a randomly distributed and unbiased sample size, the sample mean and population mean will be equal, according to the central limit theorem. Therefore, the mean of the sample means will always approximate the population mean.

8 0

3 years ago

How many square feet is 15' x 19'

Nuetrik [128]

285 is square feet lf 15x19 baselcy 15 • 19 = 285

3 0

3 years ago

The expression 12xyz-45 is a a. monomial b. binomial c. trinomial d. constant

ANEK [815]

Answer:

Hey there!

12xyz-45 is a binomial. A binomial contains two terms.

Let me know if this helps :)

4 0

3 years ago

A tablet weighs 1.23 pounds. What is its weight written as a mixed number? Plus workings

Studentka2010 [4]

1 23/100 ...............................................

6 0

3 years ago

An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click

Sergeeva-Olga [200]

Answer:

A. $\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

B. $\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

Step-by-step explanation:

Given that:

An investment of Amount = $8000

earns at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A) Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

$A = Pe^{rt}$

where;

$A = (8000) \ e ^{0.7t}$

The instantaneous rate of change = $\dfrac{dA}{dt}$

$\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )$

$= 8000 \dfrac{d}{dt}e^{0.07 \ t}$

$\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}$

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 2 years; the instantaneous rate of change is:

$\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}$

$\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

$12000 = (8000)e^{0.07 \ t}$

$\dfrac{12000 }{8000}= e^{0.07 \ t}$

$1.5= e^{0.07 \ t}$

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 5.79

$\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}$

$\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

7 0

3 years ago