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2 years ago

Case Study 2

Mathematics

1 answer:

ICE Princess25 [194]2 years ago

4 0

Answer:

good for them i think

Step-by-step explanation:

this doesnt make any sense

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36 is 120% of what number

nadezda [96]

Let, the number = x
Then,x * 120% = 36
x * 1.20 = 36 [ 120% = 1.20 ]
x = 36 / 1.20
x = 30

In short, Your Answer would be 30

Hope this helps!

6 0

3 years ago

Read 2 more answers

to determine whether a number is written in scientific notation what test can you apply to the first factor and what test can yo

pashok25 [27]

Answer:

Step-by-step explanation:

The first term must have one number before decimal point

the second term (factor) must be 10 having some power.

example;

5.0 ×10³

7.002×10^41

6 0

3 years ago

14 is 28% of what number

Juliette [100K]

14 is 28% of the number 50.

5 0

3 years ago

Kristen invests $5,745 in a bank. The bank pays 6.5% interest compounded monthly. How long must she leave the money in the bank

s344n2d4d5 [400]

formula is $A=P(1-\frac{r}{n})^{nt}$

where A=final amount

P=principal

r=interest rate in decimal

n=number of times per year it is compounded

t=time in years

we want to find where

A=2P

and P=5745

and r=6.5%=0.065

n=monthly=12

remember that $ln(x^a)=a(ln(x))$

also that $(a^b)^c=a^{bc}$

$2(5745)=5745(1+\frac{0.065}{12})^{12t}$ , solving for t

divide both sides by 5745 to simplify things a bit

$2=(1+\frac{0.065}{12})^{12t}$ I'd rather not simplify this because it give us a decimals and those aren't exact, if we combine, we get 12.065/12 for inside parenthases

$2=(\frac{12.065}{12})^{12t}$

take ln of both sides

$ln(2)=ln((\frac{12.065}{12})^{12t})$

$ln(2)=(12t)ln(\frac{12.065}{12})$

divide both sides by $12ln(\frac{12.065}{12})$

$\frac{ln(2)}{12ln(\frac{12.065}{12})}=t$

using our calculator, t≈10.6927

so rounded, we get 10.7 years

3 0

3 years ago

The endpoints of a line segment are (-2, -4) and (6, 2) What are the coordinates of the midpoint? 1. (2, 3) 2. (4,3) 3. (4, -2)

marishachu [46]

It should be 4. (2,-1)

8 0

3 years ago