Evaluate the expression. (6-2)[(2+3)+4 A) 9 B) 13 C) 24 D)36

Does anyone know how to do this? I’m lost

Nataly [62]

Answer: B

Step-by-step explanation:

To find the inverse function, you switch x with y and y with x. Then, you solve for y.

$y=(x-1)^{2} -4$ [switch x with y, and y with x]

$x=(y-1)^2-4$ [add both sides by 4]

$x+4=(y-1)^2$ [square root both sides]

$\sqrt{x+4} =y-1$ [add both sides by 1]

$\sqrt{x+4} +1=y$

$f^-^1(x)=1+\sqrt{x+4}$

Now that we know the inverse function, we can actually tell that the answer is B. We know this because of the restriction that f(x) is x≤1. This makes it positive, and B the answer.

7 0

3 years ago

Write as a product of two polynomials. (x-y)^2-a(y-x)

11Alexandr11 [23.1K]

Answer:

(x-y) (a+x-y)

Step-by-step explanation:

(y-x)=-(x-y)

-a(y-x) = a(x-y)

(x-y)^2 = (x-y)(x-y)

(x-y)(a + x - y)

3 0

4 years ago

-3√-20 written as a complex number

galben [10]

Answer: 3√-20 = 3√20√-1 = (3√20)i

Step-by-step explanation:

√-20 can be written √20√-1 since

(√20√-1)^2 is indeed -20.

3 0

3 years ago

An electrician charges $40 for each hour he works plus a $125 service charge. The total charge for a recent job was $1,205. Whic

maks197457 [2]

Answer:

Answer is D.

Step-by-step explanation:

40h is what you start out with since it costs 40 ever hour. then you add the service charge which makes it 40h + 125, and then since the total charge is 1205, that would complete the equation.

5 0

3 years ago

Read 2 more answers

A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o

NeX [460]

Answer:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

$\hat p=\frac{115}{180}=0.639$ estimated proportion of adults with the medication was effective

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is less than 0.65.:

Null hypothesis: $p \geq 0.65$

Alternative hypothesis: $p < 0.65$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about 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 provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

3 0

3 years ago