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2 years ago

The number of points scored by Reuben in the first five basketball games of a season are shown below.

Mathematics

1 answer:

jolli1 [7]2 years ago

8 0

Answer:

The correct option is (D).

Step-by-step explanation:

The data for the number of points scored by Reuben in the first five basketball games of a season are:

A = {23, 15, 28, 19, 25}

And the data for the number of points scored by Reuben in the next five games is:

B = {23, 30, 15, 22, 26}

For the first data there are only 5 observations to draw conclusions about the spread of the data and the peak of the distribution curve.

But after the next five games played by Reuben, there are 10 observations to draw conclusions.

The spread of the data is dependent upon the sample size.

Larger the sample size the smaller is the spread. And hence the distribution curve will be more peaked.

And smaller the sample size the larger is the spread. And hence the distribution curve will be less peaked.

So, after the next five games the distribution of Reuben's score would become more peaked and less widely spread.

The correct option is (D).

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Pre-Calc: Find all the zeros of the function.

uysha [10]

The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)