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3 years ago

If f(x)= x-3/x and g(x)= 5x-3, what is the domain of (f•g)(x)

1 answer:

4 0

$f(x)=\dfrac{x-3}{x},\ D_f:x\neq0\\\\g(x)=5x-3,\ D_g=\mathbb{R}\\\\(f\cdot g)(x)=f(x)\cdot g(x)\\\\(f\cdot g)(x)=\dfrac{x-3}{x}\cdot(5x-3)=\dfrac{(x-3)(5x-3)}{x}\\\\D_{(f\cdot g)}=D_f$

Answer: The domain of (f•g)(x) is all real numbers except for 0

$D=\mathbb{R}-\{0\}$

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PLS HURRY WILL GIVE BRAINLIEST

marta [7]

Answer:

C. y= 6x-5

Step-by-step explanation:

Given Information :

Slope (m) = 6

y-intercept (b) = -5

Equation of a line :

y=mx+b

Where :

m = slope

b = y-intercept

So , with the given the information , the equation is :

$y = 6x - 5$

7 0

3 years ago

In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom

coldgirl [10]

Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they would watch one of the television shows not differs from 0.5 or 50% .

Step-by-step explanation:

1) Data given and notation n

n=106 represent the random sample taken

X=48 represent the people who says that they would watch one of the television shows.

$\hat p=\frac{48}{106}=0.453$ estimated proportion of people who says that they would watch one of the television shows.

$p_o=0.5$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that 50% of people who says that they would watch one of the television shows.:

Null hypothesis: $p=0.5$

Alternative hypothesis: $p \neq 0.5$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968$

4) Statistical decision

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level is not provided, but we can assume $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So based on the p value obtained and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they would watch one of the television shows not differs from 0.5 or 50% .

6 0

3 years ago

Can someone help me in this plssss I really need it

kompoz [17]

Answer:

$\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}$

Step-by-step explanation:

Your expression is

$\sqrt [3] {-54x^{5}y^{6}}$

Here's how I would simplify it.

$\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}$

$\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}$