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3 years ago

Is the function represented by the table non-linear?

1 answer:

7 0

No, because it has a constant rate of change. ... If the rate of change is not constant, then the function cannot be linear

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BOSE

Lina20 [59]

Answer:

a = 1 ; b = -8 ; c = -5

Step-by-step explanation:

f(p) = p² - 8p -5

a = co-efficient of p² = 1

b = co-efficient of p = -8

c = constant = -5

8 0

2 years ago

The side length of a square is represented by the expression 2x + 5. Which expression represents the difference between the area

musickatia [10]

Answer:

$4x^2 + 22x + 15$

Step-by-step explanation:

The side length of a square is represented by the expression 2x + 5.

The area of a square is given as:

$A = a^2$

where a = length of side of the square

The area of the square is therefore:

$A = (2x + 5)^2\\\\A = (2x + 5)(2x + 5)\\\\A = 4x^2 + 20x + 25$

The perimeter of a square is given as:

$P = 4a$

The perimeter of the square is therefore:

$P = 4(2x + 5) \\\\P = 8x + 10$

The difference between the area of the square and the perimeter of the square is:

$4x^2 + 30x + 25 - (8x + 10)\\\\4x^2 + 30x + 25 - 8x - 10\\\\4x^2 + 30x - 8x + 25 -10\\\\4x^2 + 22x + 15$

The expression that represents the difference between the area and the perimeter of the square is:

$4x^2 + 22x + 15$

7 0

2 years ago

The distribution of IQ scores can be modeled by a normal distribution with mean 100 and standard deviation 15.

evablogger [386]

Answer:

4.4% of the population with IQ between 120 and 125.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 15

We are given that the distribution of IQ scores is a bell shaped distribution that is a normal distribution.

a) Let X be a person's IQ score.

Then, density functions for IQ scores is given by:

$P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{z^2}{2}}\\\\\text{where,}\\\\z = \frac{x-\mu}{\sigma}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-100)^2}{450}}$

b) P(population with IQ between 120 and 125.)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

$P(120 \leq x \leq 125) = P(\displaystyle\frac{120 - 100}{15} \leq z \leq \displaystyle\frac{125-100}{15}) = P(1.33 \leq z \leq 1.66)\\\\= P(z \leq 1.66) - P(z < 1.33)\\= 0.952 - 0.908 = 0.044 = 4.4\%$