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3 years ago

True or false: x/3 is equivalent to 1/3 x. Explain your answer.

Mathematics

1 answer:

GalinKa [24]3 years ago

7 0

True
No matter what you fill in it multiples the other fraction too so that they are always equal

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Use DESMOS: What is the equation

Masja [62]

y=mx+c

I picked the points (0, -4) and (-2,-10) to find the gradient of the line

M= rise/run

= 6/2

Y-intercept is at - 4

Y= 3x - 4 is equation of the line given

Reciprocal of 3 is 1/3

(m1 x m2 = - 1)

So, to make - 1, our second gradient should be negative.

Gradient of the perpendicular line will be - 1/3

Y= - 1/3x + c

Substitute value of x & y into the equation

(-6,6) means x = - 6, y = 6

y = - 1/3x +c

6 = - 1/3 x - 6 + c

6 = 2 + c

-2

4 = c

Thus, final equation we're after will be:

y= - 1/3x + 4

Hope this helps!

6 0

2 years 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 Two marbles are withdrawn randomly.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

topjm [15]

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We can assume that both marbles are withdrawn at the same time.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 The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .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.

Probability of winning: $\frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}$

Probability of losing $\frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}$

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing = $1.10\cdot \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}$

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing = $1.10^2\cdot \frac{4}{9}+\frac{5}{9}=\frac{82}{75}$

We will use the following formula

$Var(X) = E(X^2)-E(X)^2$

Thus

Var(X) = $\frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}$

7 0

4 years ago

Keith will run at most 39 miles this week. So far, he has run 21 miles. What are the possible numbers of additional miles he wil

Luda [366]

Uhhh 18?? 9+9, 10+8, 7+11, 6+12, 5+13, 4+14, 3+15, 2+16, 1+17

3 0

3 years ago

I'm not sure what they mean by "relations that represents a function". I don't understand this.

sergeinik [125]

Means, for every 1 input, there is 1 coresponding output
normally x is input and y is output in form (x,y)
so just look for the option(s) that has/have every first number repeat with only 1 2nd number

A. -7 repeats with 5 and -3, not a function

B. no repeats of first number (6,-8) and (-5,-8) are fine because first numbers don't repeat, das is funciton

C. no repeats, function

D. no repeats, function

answer is B,C,D

4 0

3 years ago

The graph shows that f(x)=(one-third) Superscript x is translated horizontally and vertically to get the function (one-third) Su

harina [27]

It is the number 3.

6 0

3 years ago