Decide whether each equation is true for all, one, or no values of x. -2x + 4 = -3x + 4

Step-by-step explanation: When finding the MAD, you need to find the mean of the data, which they said is 10. The next step is to subtract numbers from the data from the mean, which is 10 for all three dot plots. For example, for Jason's team, we need to subtract all the 7's, 8's, 11's, 12's, and 13's, because those are the only parts of the plot with data. Let's try subtracting 7 - 10. The answer is -3, but we need to make it positive, because that's how data works, so it's actually 3. You keep doing this for all three plots, which is provided below:

Jason's Team: 3, 3, 2, 2, 2, 1, 2, 3, 3, 3.

Hannah's Team: 3, 2, 2, 1, 0, 0, 1, 1, 3, 3.

Dion's Team: 1, 1, 1, 1, 0, 0, 0, 0, 1, 3.

Now, the next step is adding all these values and dividing it by ten, since that's how many numbers there are for the data.

Jason's Team MAD: 2.4.

Hannah's Team MAD: 1.6.

Dion's Team MAD: 0.8

As you can see, Jason's Team has the greatest MAD, which is your answer.

Let me know if this is right! :)

8 0

3 years ago

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0

4 years ago