All categories

4 years ago

What is the GCF of the 48and16

Mathematics

2 answers:

zavuch27 [327]4 years ago

6 0

The GCF would be 16.

rjkz [21]4 years ago

4 0

The greatest common factor of 48 and 16 should be 16

You might be interested in

A cyclist rides his bike at a speed of 18 miles per hour. What is this speed in kilometers per hour? How many kilometers will th

Sonja [21]

Well, if 1 mile = 1.6 kilometers and he's traveling at 18mph, then do 18 x 1.6 which = 28.8.

So, this is his current speed in kph

3 x 28.8 = 86.4‬

They will have traveled 86.4‬ kilometers

(I haven't worked with this kind of math in a while so correct me if I'm wrong!)

4 0

3 years ago

A person places $7320 in an investment account earning an annual rate of 8.2%,

klasskru [66]

Answer:

The amount of money after 12 years is $19581.99 to the nearest cents

Step-by-step explanation:

The formula of the compound continuously interest is V = P $e^{rt}$ , where

V is the value of the account in t years

P is the principal initially invested

e is the base of a natural logarithm

r is the rate of interest in decimal

∵ A person places $7320 in an investment account

∴ P = 7320

∵ The account earning an annual rate of 8.2%, compounded continuously

∴ r = 8.2% ⇒ divide it by 100 to change it to decimal

∴ r = 8.2 ÷ 100 = 0.082

∵ The time is 12 years

∴ t = 12

→ Substitute these values in the formula above to find V

∵ V = 7320 $e^{0.082(12)}$

∴ V = 19581.99121 dollars

→ Round it to the nearest cents ⇒ 2 d.p

∴ V = 19581.99 dollars

∴ The amount of money after 12 years is $19581.99 to the nearest cents.

8 0

3 years ago

I need help right now please

artcher [175]

Your answer is c. i know this because this was the exact same question i did yesterday and i got 100%

6 0

3 years ago

A sample of 25 undergraduates reported the following dollar amounts of entertainment expenses last year:

marin [14]

Answer:

$Range = 85$

$\sigma = 28.71$

$Interval = [666.78, 781.62]$

Step-by-step explanation:

Given

The data for 25 undergraduates

Solving (a): Range and Standard deviation

The range is:

$Range = Highest - Least$

From the dataset:

$Highest = 772$

$Least = 687$

So:

$Range = Highest - Least$

$Range = 772-687$

$Range = 85$

The standard deviation is:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

First, calculate the mean

$\bar x = \frac{769 +691 +............+715}{25}$

$\bar x = \frac{18105}{25}$

$\bar x = 724.2$

So, the standard deviation is:

$\sigma = \sqrt{\frac{(769-724.2)^2 +(691-724.2)^2 +(699-724.2)^2 +(730-724.2)^2 +............+(715-724.2)^2}{25}}$

$\sigma = \sqrt{\frac{20604}{25}}$

$\sigma = \sqrt{824.16}$

$\sigma = 28.71$

Solving (b): The interval of the 95% of the observation.

Using the emperical rule, we have:

$Interval = [\bar x - 2*\sigma, \bar x+ 2*\sigma]$

$Interval = [724.2 - 2*28.71, 724.2 + 2*28.71]$

$Interval = [666.78, 781.62]$

4 0

3 years ago

Karen is working on her math homework. She solves the equation

Gelneren [198K]

Disagree.
b/8 = 56; multiply both sides by 8 to solve for b, and you get b = 448

5 0

3 years ago