6th grade math I mark as brainliest

Maru [420]

The true statement about these graph is that: B. the graph of the original function (y = 8/5x + 4) is perpendicular to the graph of the new function (y =- 5/8x + 8).

<h3>What is a graph?</h3>

A graph can be defined as a type of chart that's commonly used to for the graphical representation of data on both the horizontal and vertical lines of a Cartesian coordinate, which are typically known as the x-axis and y-axis respectively.

<h3>How to interpret the graph?</h3>

Mathematically, the standard form of the equation of a straight line on a graph is given by;

y = mx + c

<u>Where:</u>

x and y are the points.

m is the slope.

c is the intercept.

<h3>The condition for perpendicularity.</h3>

In Mathematics, the condition for perpendicularity of two (2) lines is as follows:

m₁ × m₂ = -1

8/5 × (-5/8) = -1

Based on the graph (see attachment) of the original function and new function, we can infer and logically deduce that the true statement about their graph is that the graph of the original function (y = 8/5x + 4) is perpendicular to the graph of the new function (y =- 5/8x + 8).

Read more on perpendicularity here: brainly.com/question/1202004

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Complete Question:

Suppose the following function is graphed. y = 8/5x + 4 On the same grid, a new function is graphed. The new function is represented by the following equation. y =- 5/8x + 8. Which of the following statements about these graphs is true?

A. The graphs intersect at (0,8).

B. The graph of the original function is perpendicular to the graph of the new function.

C. The graph of the original function is parallel to the graph of the new function.

D. The graphs intersect at (0,4).

3 0

2 years ago

Two fair dice are rolled. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any di

kvasek [131]

Answer:

a)

P(X = x₀, Y = 2x₀) = 1/36

P(X = x₀, Y = k) = 1/18 for k between x₀+1 and 2x₀-1 inclusive

Every other event has probability 0. x₀ is any number between 1 and 6 inclusive.

b)

P(X = x₀, Y = x₀) = x₀/36

P(X = x₀, Y = k) = 1/36 for k between x₀+1 and 6 inclusive.

x₀ is between 1 and 6 inclusive. Every other event has probability 0.

c)

P(X = x₀, Y = x₀) = 1/36

P(X = x₀, Y = k) = 1/18 with k between x₀+1 and 6 inclusive

x₀ between 1 and 6 inclusive. Any other event has probability 0.

Step-by-step explanation:

Note that there are 36 possible results for the dice

a)

P(X = 1, Y = 2)

This is obtained only when both dices are 1, hence its probability is 1/36

P(X = 1, Y = k) = 0 (k > 1)

because if the largest value of the dice is 1, then both dices are 1

P(X = 2, Y = 3)

one dice is 2, the other one is 3, hence there are 2 possibilities and the probability is 2/36 = 1/18

P(X = 2, Y = 4)

This happens only if both dices are 2, hence the probability is 1/36.

P(X = 2, Y = k) = 0 (k > 2)

same argument of above. If the largest dice is 2, then the sum is either 3 or 4.

P(X = 3, Y = 4), P(X = 3, Y = 5)

in both given events we need one dice to be 3 and the other dice to be 1 for the first event and 2 for the second event. In both cases, there are only 2 favourable cases, hence the probability of the event is 2/36 = 1/18

P(X = 3, Y = 6)

This event happens only when both dices are 3, hence the probability is 1/36

This should show a pattern. As long as x₀ is between 1 and 6, if y₀ is between x+1 and 2x-1, then the probability P(X = x₀, Y = y₀) is 1/18 (either first dice is x₀, second dice is y₀-x₀ or first dice is x₀ and second dice is y₀ - x₀), also P(X = x₀, Y = 2x₀) = 1/36 (both dices are x₀). Every other event has probability 0.

b) We can separate them using conditional probability and the fact that both dices results are independent with each other.

P(X = x₀, Y = y₀) = P(X = x₀) * P(Y = y₀ | X = x₀)

P(X = x₀) = 1/6 for any value x₀ between 1 and 6.

If y₀ is x₀, this means that the first dice has the largest value, so the second dice is between 1 and x₀, and the probability of this event is x₀/6 (x₀ favourable cases over 6 possible ones).

If y₀ is not x₀, then it should be higher (otherwise the event would be impossible and it would have probability 0). As long as y₀ is between 2 and 6, the probability of this event is 1/6.

Thus

P(X = x₀, Y = x₀) = 1/6 * x₀/6 = x₀/36

P(X = x₀, Y = x₀ + k) = (1/6)² = 1/36 (k > 0)

Every other probability is 0

c)

P(X = x₀, Y = x₀) = 1/36 (because both dices are equal to x₀ in this event)

P(X = x₀, Y = x₀+k) = 2/36 = 1/18 (here k > 0. One possibility is the first dice is x₀ and the second one is x₀+k, and the remaining possibility is the first dice is x₀+k and the second dice is x₀)

Evert other event has probability 0.

4 0

3 years ago

The circle below is centered at the point (3,1) and has a radius of length 2. What is its equation?

Paul [167]

It is letter B.
Equation of a circle is in the form:
(x-h)^2 + (y-k)^2 = r^2
where (h,k) is the center of the circle with radius r.

3 0

3 years ago

Read 2 more answers

Which describes the cost to produce one item?

Vlada [557]

The answer is either B or D