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3 years ago

Use the functions h(x) = 2x + 5 and t(x) = 7x − 6 to complete the function operations listed below.

1 answer:

5 0

A. (m+n)(x)=
5x+4+6x-9=
11x-5

B. (mn)(x)=
(5x+4)(6x-9)=
30x^2+24x-45x-36=
30x^2-21x-36

C. basically sub n(x) for x in m(x)
m(n(x))=
5(6x-9)+4=
30x-45+4=
30x-41

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to solve the system of equations below, grace isolated the variable y in the first equation and then substituted into the second

Marizza181 [45]

Answer:

The resulting equation is

x^2/4+16x^2/9=1

Step-by-step explanation:

The given equations are:

3y=12x eq(1)

x^2/4+y^2/9=1 eq(2)

We need to isolate variable y in equation 1

Divide both sides of the equation with 3

3y/3 = 12x/3

y = 4x

Now, substitute the value of y=4x in second equation

x^2/4+y^2/9=1

x^2/4 + (4x)^2/9 = 1

The resulting equation is

x^2/4+16x^2/9=1

8 0

4 years ago

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What is the equation of the line in slope-intercept form?

Bingel [31]

Y= Mx+b

In the equation of a straight line,the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis).

7 0

4 years ago

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Need an equation that makes a parabola from soccer ball over soccer dummy and into the garbage can

dezoksy [38]

Answer:

The position of David Beckham is at (2,0) and the position of the garbage can is (38,0). Note that these two points would be the roots of the parabola we are going to draw.

Now since we have the roots, we can construct the equation of the parabola as

$-a(x-2)(x-38)$

where a is a constant.

We choose a = 0.05 and get

$-0.05(x-2)(x-38)$

4 0

2 years ago

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

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

Formula:

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

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

3 years ago

Basic Computation: Testing M1 2 M2 A random sample of 49 measurements from a population with population standard deviation 3 had

Elenna [48]

Answer:

The sample statistics follows a standard normal distribution since the sample size are large enough.

Step-by-step explanation:

Given that:

First population:

Sample size $n_1$ = 49

Population standard deviation $\sigma_1$ = 3

Sample mean $\overline x _1$ = 10

Second population:

Sample size $n_2$ = 64

Population standard deviation $\sigma_2$ = 4

Sample mean $\overline x_2$ = 12

The sample statistics follows a standard normal distribution since the sample size are large enough.

The null and alternative hypotheses can be computed as:

$\mathbf{H_o:\mu_1=\mu_2}$

$\mathbf{H_1:\mu_1\ne\mu_2}$

Level of significance = 0.01

Using the Z-test statistics;

$Z = \dfrac{\overline x_1 - \overline x_2}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}$

$Z = \dfrac{10- 12}{\sqrt{\dfrac{3^2}{49} + \dfrac{4^2}{64}}}$

$Z = \dfrac{-2}{\sqrt{\dfrac{9}{49} + \dfrac{16}{64}}}$

$Z = \dfrac{-2}{\sqrt{0.18367 +0.25}}$

$Z = \dfrac{-2}{\sqrt{0.43367}}$

$Z = \dfrac{-2}{0.658536}$

Z = - 3.037

Z $\simeq$ - 3.04

The P-value = 2P (z < -3.04)

From the z tables

= 2 × (0.00118)

= 0.00236

Thus, since P-value is less than the level of significance, we fail to reject the null hypotheses $H_o$

5 0

3 years ago