-1/3(2p-9) what does p equal

Let f(x) = 3x - 6 and g(x) = x - 2. Find f/g and its domain

Neko [114]

F(x) = 3x-6
g(x) = x-2

so f/g = 3(x-2)/(x-2) = 3

hope this will help you

7 0

2 years ago

a carpenter is constructing a staircase in a house the distance from the first floor to the basement is 10.3 feet the staircase

Olegator [25]

Check the picture below.

make sure your calculator is in Degree mode.

7 0

2 years ago

Wesleys mother is 57 years old. her age is three years more than twice edwins age. what is wesleys age?

natita [175]

Answer: Edwin is likely a typo for Wesley.

Step by step:

Wmom=57

Wmom=3+2e

57=3+2e

54=2e

27=e

Edwin is 27 years old.

Aside from a typo, one can speculate on Wesley's age. He's likely between 20 and 39 years old based on his mother's age. If you want to delve into societal norms, Wesley is likely to be a name from a white middle to upper class parentage, so it's less likely he was had when his mother was a teen, or even in her early twenties because college, so move that age zone to 20 to 34. Then consider that having children later than 35 is consideration for a "risky" pregnancy and most women are urged to aim for before 33, but most upper middle class white women will be pressured to baby before 30. So, adjust Wesley's age zone to 27 to 34. And then...idk. But in all honesty I think Edwin is meant to be Wesley, because it's a math problem not a sociology one.

7 0

3 years ago

O esboço do gráfico seguinte representa uma função. Descreva o conjunto domínio D(f) e o conjunto imagem Im(f) desse função:

Harlamova29_29 [7]

Answer:

xonjustő façţor si de amor ěqual funçaø de amor!! to ďęčimăĺ..

Step-by-step explanation:

dea de gracias!!

5 0

2 years ago

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating

Natali [406]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565$

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

6 0

2 years ago