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Alona [7]
3 years ago
10

How can I solve this problems?

Mathematics
1 answer:
nadezda [96]3 years ago
7 0
1)
Julia = Sj = 30,500 + 500n
Aysha = Sa = 26,000 + 1000n

when will
Sj = Sa
30,500 + 500n = 26,000 + 1000n
30,500 - 26,000 = 1000n - 500n
4,500 = 1000n - 500n
4,500 = 500n
n = 9
so they will have same salary after 9 years

2) X + Y = 8000
5%X + 3.25%Y = 312.5

X = 8000 - Y
0.05 ( 8000 - Y ) + 0.0325Y = 312.5
400 - 0.05Y + 0.0325Y = 312.5
400 - 0.0275Y = 312.5
400 - 312.5 = 0.0275Y
87.5 = 0.0275Y
Y = 3,181.81
X = 8000 - Y = 8000 - 3,181.81 = 4,818.19

3) X + Y = 3050
8%X + 7.5%Y = 234

X = 3050 - Y
0.08(3050 - Y) + 0.075Y = 234
244 - 0.08Y + 0.075Y = 234
244 - 234 = 0.025Y
Y = 400
X = 3050 - Y = 3050 - 400 = 2650



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maks197457 [2]

Answer:

1.

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2

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3

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8 0
3 years ago
What is the solution to this equation <br> 4x+x-15+3-8x=13
Fynjy0 [20]
5x -15 + 3 -8x = 13
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3 0
3 years ago
The population of rabbits on an island is growing exponentially. In the year 1994, the population of rabbits was 9600, and by 20
drek231 [11]

Answer:

49243

Step-by-step explanation:

Given that the population of rabbits on an island is growing exponentially.

Let the population, P=P_0e^{bt}

where, P_0 and b are constants, t=(Current year -1994) is the time in years from 1994.

In 1994, t=0, the population of rabbit, P=9600, so

9600=P_0e^{b\times 0}

So, P_0=9600

and in 2000, t=2000-1994=6 years and population of the rabbit, P=18400

18400=9600 \times e^{b\times 6} \\\\\frac{18400}{9600}=e^{b\times 6} \\\\

\ln(23/12}=6b \\\\

b = \frac{\ln{1.92}}{6} \\\\

b=0.109

On putting the value of P_0 and b, the population of the rabbit after t years from 1994 is

P=9600 \times e^{0.109\times t}

In 2009, t= 2009-1994=15 years,

So, the population of the rabbit in 2009

P=9600 \times e^{0.109\times 15}=49243

Hence, the population of the rabbit in 2009 is 49243.

7 0
3 years ago
Use quadratics---------------
Elza [17]
A square's area is the square of the sides 

square area= side²
side=\sqrt{96x^5} &#10;
side=\sqrt{2(2)(2)(2)(3)x^2(x)}
side=4x^2 \sqrt{3x}

OR we can divide it 
    |_ 96x^5
 x²|_ 96x^3
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5 0
3 years ago
The state highway department is studying traffic patterns on one of the busiest highways in the state. As part of the study, the
larisa86 [58]

Answer:

The need to sample 76 days.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many days do they need to sample

We need to sample n days.

n is found when M = 300.

We have that \sigma = 1010

So

M = z*\frac{\sigma}{\sqrt{n}}

300 = 2.575*\frac{1010}{\sqrt{n}}

300\sqrt{n} = 2.575*1010

\sqrt{n} = \frac{2.575*1010}{300}

(\sqrt{n})^{2} = (\frac{2.575*1010}{300})^{2}

n = 75.1

Rounding up

The need to sample 76 days.

3 0
3 years ago
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