Ask question

Ask question

All categories

harkovskaia [24]

3 years ago

15

1) Sandra saves 12% of her salary for retirement. This year her salary was $3000 more than in the previous year, and she saved $

4,200. What was her salary in the previous year? i want to know the equation form
the topic is math 8 grade writing equations

2 answers:

OverLord2011 [107]3 years ago

8 0

Hello there.
<span>
1) Sandra saves 12% of her salary for retirement. This year her salary was $3000 more than in the previous year, and she saved $4,200. What was her salary in the previous year?
</span>
32,000

svp [43]3 years ago

3 0

Given:
savings 12% of her salary.
savings $4,200
This year's salary: x + 3,000
Last year's salary: x

4,200 represents 12% of her salary.
4,200 / 12% = 35,000 is her salary this year.

x + 3,000 = 35,000
x = 35,000 - 3,000
x = 32,000 her salary last year.

You might be interested in

erica [24]

That is true because the square root of 32 is about 5.66, so 4.65 IS less than it.

3 0

3 years ago

PLZ HELP FAST!!! (test)

Olenka [21]

I can’t see the tea really good

6 0

3 years ago

Read 2 more answers

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

Scott is playing a game in which he rolls a number cube with faces numbered 1through 6 and spins the spinner shown below one tim

defon

Solution:

The total number of possiblities rolling a number cube with faces numbered 1 to 6 is;

And the total number of possibilities spining the spinner is;

The number of possibility where outcome on the cube is greater than 2 is;

$4$

And the number on the spinner less than 9 is;

$2$

Hence, the unique combination is;

$4\times2=8$

CORRECT OPTION: A

7 0

1 year ago

Juans three math quizzes this week took him 1/3?4/6?and1/5 hour to complete. List all the fraction from least to greatest

Alex

Answer:

$\frac{1}{5}$ , $\frac{1}{3}$ , $\frac{4}{6}$ .

Step-by-step explanation:

Given fractions $\frac{1}{3}$ , $\frac{4}{6}$ , $\frac{1}{5}$ .

We need to list them from least to greatest.

In order to arrange them from least to greatest, we need to find the least common denominator of $\frac{1}{3}$ , $\frac{4}{6}$ , $\frac{1}{5}$ .

We have 3, 6 and 5 in denominators.

Least common multiple of 3, 6 and 5 is = 30, because 30 is least number that can be divided by all three number 3, 6 and 5.

Let us covert each denominator as 30.

Multiplying first fraction $\frac{1}{3}$ by 10 in top and bottom, we get

$\frac{1}{3} = \frac{1\times10}{3\times10}=\frac{10}{30}$

Multiplying first fraction $\frac{4}{6}$ by 5 in top and bottom, we get

$\frac{4}{6} = \frac{4\times5}{6\times5}=\frac{20}{30}$

Multiplying first fraction $\frac{1}{5}$ by 6 in top and bottom, we get

$\frac{1}{5} = \frac{1\times6}{5\times6}=\frac{6}{30}.$

Now, we can check $\frac{10}{30}, \frac{20}{30} \ and \ \frac{6}{30}.$

$\frac{6}{30}$ is the smallest, $\frac{10}{30}$ is greater and $\frac{20}{30}$ is the greatest.

Therefore, we can arrange fractions $\frac{6}{30}, \frac{10}{30} \ and \ \frac{20}{30}.$

Writing original fractions in place of equivalent fractions, we can write

$\frac{1}{5}$ , $\frac{1}{3}$ and $\frac{4}{6}$ .

Therefore, the order the amounts of paint from least to greatest is:

$\frac{1}{5}$ , $\frac{1}{3}$ , $\frac{4}{6}$ .

4 0

3 years ago