7* = 343/7* = 56 solve??? help me please

Find the mean of the following data set. 42, 45, 58, 63 42 47 52

ohaa [14]

Answer:

d

Step-by-step explanation:

6 0

3 years ago

The coach recorded how fast his runners ran 1 mile. Emmett ran a mile in 9.73 minutes Kate ran a mile in 10.87 minutes how many

igor_vitrenko [27]

Complete Question

The coach recorded how fast his runners ran 1 mile. Emmett ran a mile in 9.73 minutes, Kate ran a mile in 10.87 minutes, and Rubin ran a mile in 8.46 minutes. How many more minutes did Emmett run than Robin?

a. 1.37 minutes

b. 1.27 minutes

c. 2.27 minutes

d. 1.27 seconds

Answer:

1.27 minutes

Step-by-step explanation:

Emmett ran a mile in 9.73 minutes

Kate ran a mile in 10.87 minutes

Rubin ran a mile in 8.46 minutes.

How many more minutes did Emmett run than Robin?

This is calculated as:

Number of minutes Emmett ran - Number of minutes Rubin ran

= 9.73 minute - 8.46 minutes

= 1.27 minutes

Option b is correct

5 0

3 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

4 years ago

Identify the 34th term of the arithmetic sequence 2, 7, 12

torisob [31]

The equation is AN=A1+D(N-1) A1 being the first number (2) D being the difference (in this case 5) and N being the nth term to find (34) so here you have AN=2+5(34-1) then AN=2+5(33). the answer here is 167.

5 0

3 years ago

Solve. m - 15 = 20 m =

Fittoniya [83]

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