PLEASE HELP ME WITH BOTH 9 AND 10!!!

Which of the following values of x will result in a true statement when substituted into the given equation?

Andrej [43]

Answer:

B.X=-1

Step-by-step explanation:

-3 (X-2)=9

-3X+6=9

-3X=9-6

-3X=3

X=3/-3

X=-1

7 0

3 years ago

Donald can read 3/4 of a book in 1 1/2 hours.

Inessa [10]

Answer:

Thats fun but i want to help! what the question?

Step-by-step explanation:

8 0

3 years ago

The length of the track at talladega superspeedway is 2.66 miles. If the race is 188 laps long , how many kilometers do the driv

kupik [55]

Answer:

785

Step-by-step explanation:

Multiply the length of the track by the amount of laps to get length in miles

188 * 2.6 = 487.8

then covert the miles to kilometers to get 785 kilometers

4 0

4 years ago

Please help asap! Factor the polynomial

Afina-wow [57]

C is the answer but your hot add me on snap sosa.adrian

5 0

4 years ago

In a survey, 32% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result

adoni [48]

Answer:

$z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932$

$p_v =P(z$

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 pet owners contacted by telephone is NOT significantly lower than 0.32

Step-by-step explanation:

Data given and notation

n=210 represent the random sample taken

X=61 represent the number of pet owners contacted by telephone

$\hat p=\frac{61}{210}=0.29$ estimated proportion of pet owners contacted by telephone

$p_o=0.32$ 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 the true proportions is lower than 0.32.:

Null hypothesis: $p \geq 0.32$

Alternative hypothesis: $p < 0.32$

When we conduct a proportion test we need to use the z statistic, 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:

$z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932$

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 pet owners contacted by telephone is NOT significantly lower than 0.32

3 0

3 years ago