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2 years ago

The area of a rectangular field is 2275 m2. The field is enclosed by 200 m of fence. What are the dimensions of the field?

1 answer:

4 0

So it is enclosed by 200 m of fence
perimiter=200 m
it is a rectangle to
p=legnth+legnth+width+width or
p=l+l+w+w or
p=2l+2w or
p=2(l+w)
so
p=200
200=2(l+w)
divide both sides by 2
100=l+w
area=l times w
area=2275
lw=2275
l+w=100
combine and solve
l+w=100
subtract w
l=100-w
subisutute 100-w for l in other eqution
(100-w)(w)=2275
distribute
-w^2+100w=2275
add (w^2-100w) to both sides
0=w^2-100w+2275
factor
find what 2 numbers add up to -100 and multiply to get 2275
guess (or factor 2275 and find factors that add up to -100)
figure out that they are -65 and -35
0=(w-65)(w-35)
set each to zero
0=w-65
0=w-35

solve for w
w=65 or 35
65>35 so
65 m=legnth
35 m=width

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1 year ago

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Which rational number is one tick mark to the right of Negative 6 Over 4 on the number line below?

murzikaleks [220]

Answer:

The number line is missing, but as we are know that the number marked in the number line is -6/4, i will guess that the ticks are spearated by fourts (the distance between each tick is 1/4).

Now, for the number at the right of -6/4, we should add the distance for one tick, this means that the number at the right is:

-6/4 + 1/4 = -5/4.

Now i will give some other examples:

Now, if the distance between ticks is 2/4, then the number at the right will be:

-6/4 + 2/4 = -4/4 = -1

Now, if the distance between ticks is 3/4, the the number at the right will be:

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1 year ago

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if

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Answer:

a) The expected value is $\frac{-1}{15}$

b) The variance is $\frac{49}{45}$

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is $\binom{5}{2}$ . Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is $\binom{5}{1}\cdot \binom{5}{1}$ . So, we define the following probabilities.