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Shalnov [3]
3 years ago
8

Need help ? On answering number 4.

Mathematics
1 answer:
Kruka [31]3 years ago
6 0

Answer:

3.9

Step-by-step explanation:

wavy equal 3.87298

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Your anwser for letter X is 2
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Input the equation of the given line in standard form.
noname [10]

Answer:

y = 2/3x + 1/3

Step-by-step explanation:

Standard form of a line looks like y=mx+b. We already know that m is 2/3, but we don't know b. b is our y-int. With the given information, we can write this equation is point-slope form. That looks like y - y1 = m(x-x1). (x1, y1) = (1, 1) - the point given to us. So if you plug in that point to the equation, and the slope - m, it'll look like y - 1 = 2/3 (x - 1).

From here, you can just solve for y, and it'll be in standard form. I would distribute 2/3 first.

y - 1 = 2/3x - 2/3

Then, add 1 to both sides.

y = 2/3x + 1/3

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3 years ago
From the information in the graph, which team showed the least amount of improvement over last year?
Ksenya-84 [330]

Answer:

The rangers

Step-by-step explanation:

Their number of wins didn't go up. In fact their wins decreased the most.

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Barbara can walk 3200 meters in 24 minutes
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Barbra can walk 3200 meers in 24 minutes what are you asking?


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On the 1st January 2014 Carol invested some money in a bank account.
Ghella [55]

Answer:

\large \boxed{\text{\pounds 23 360.00}}

Step-by-step explanation:

The formula for the accrued amount from compound interest is

A = P \left(1 + \dfrac{r}{n}\right)^{nt}

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:  

r = 0.025

\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}

3. Summary

1 Jan 2014      P = £23 360.00

1 Jan 2015     A =    23 944.00

     Withdrawal = <u>    -1  000.00 </u>

                     P =     22 944.00

1 Jan 2016    A =    £23 517.60

5 0
3 years ago
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