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iren [92.7K]
3 years ago
8

In 2004, Cindy had $4000 in a mutual fund account. In 2005, the

Mathematics
1 answer:
gtnhenbr [62]3 years ago
4 0
6,250
1,000/4,000 = 25/100
25/100 = x/5,000
x = 1,250

5,000 + 1,250 = 6,250
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The number of students enrolled at a college is 12,000 and grows 4% each year. Complete parts (a) through (e).
Paha777 [63]

Answer:

48000

Step-by-step explanation:

12000 students

4%

1 =annually

12×4×1= 48000

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3 years ago
I NEED HELP ASAP PLS
xeze [42]

Answer:mot sire

Step-by-step explanation:

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Please help! Evaluate the following function.<br> f(x) = -x – 8 for f(-1)
Anton [14]

Answer:

x - 8

Step-by-step explanation:

Plug in -1 for x and you get x - 8

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3 years ago
Solve the system of equations algebraically.
andrew11 [14]

Answer: d. (17, 20)

Step-by-step explanation:

We have the following system of equations:

x-3y=8 (1)

5x-4y=5 (2)

Isolating x from (1):

x=\frac{8+3y}{4} (3)

Substituting (3) in (2):

5(\frac{8+3y}{4})-4y=5 (4)

Isolating y:

\frac{40}{4}+\frac{15 y}{4}-4y=5 (5)

10-5=4y-\frac{15 y}{4} (6)

y=20 (7)

Substituting (7) in (1):

x-3(20)=8 (8)

Isolating x:

x=17 (9)

Hence, the correct option is d. (17, 20)

8 0
3 years ago
39. Kate recorded the time it took six children of
scZoUnD [109]

Answer:

y=-11x+261

Step-by-step explanation:

As you can observe in the image attached, the line that best fits passes through point B and C. That means we can use those point to find the slope of such line.

m=\frac{y_{2} -y_{1} }{x_{2}-x_{1}  }

Where B(11,137) and C(12,126)

m=\frac{126-137}{12-11}=-11

So, the slope of the line that best fits is -11, approximately.

Now, we use the point-slope formula to find the equation.

y-y_{1} =m(x-x_{1} )\\y-137=-11(x-11)\\y=-11x+124+137\\y=-11x +261

Therefore, the line that best fits is y=-11x+261 approximately.

Remember, when we estimate a line for some data on a scatterplot, we are calculating an approximation, that's why we also said "approximately", because the line is an approximation where the majority of point meet.

4 0
3 years ago
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