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BartSMP [9]
3 years ago
13

You have one hour for lunch. You want to go to your favorite sub shop, which is 18 minutes from your job. You called; the averag

e wait time today is 20 minutes. How much time should you budget for your lunch trip? (Ignore time to eat)
Mathematics
2 answers:
ozzi3 years ago
7 0

Answer:

Time budget for your lunch trip is 56 minutes.

Step-by-step explanation:

Given : You have one hour for lunch. You want to go to your favorite sub shop, which is 18 minutes from your job. You called; the average wait time today is 20 minutes.

To find : How much time should you budget for your lunch trip?

Solution :

Time taken to go from sub shop to office is 18 minutes.

Same time taken from office to sub shop is 18 minutes.

Total time in travelling is 18+18=36 minutes.

The average wait time today is 20 minutes.

Total time budget for your lunch trip is 36+20=56

Therefore, time budget for your lunch trip is 56 minutes.

Shkiper50 [21]3 years ago
5 0

Answer:

About an hour, 56 mins.

Step-by-step explanation:

18+18=36

36+20=56

Voila!

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8 0
3 years ago
What does -3x + 7 + -6x t 9 equal
liberstina [14]

Answer:

-50x

Step-by-step explanation:

-6x times 9 which is -54x

-54x-3x = -57x

-54x+7

8 0
3 years ago
If mZOKL = 80°, then what is mZHKJ?<br> С<br> E<br> А4<br> N<br> H<br> K<br> M
Nookie1986 [14]
MOKL and mHKJ are vertical angles, so they’re congruent
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5 0
3 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
Glenda receives a salary of $4,792 bimonthly. Find Glenda’s earnings per year.
Andre45 [30]

Answer: $28,752

Explanation:

The<u> definition of bimonthly</u> is an event that happens once in two months. -bi represents "two", while -monthly represents "event that happens by month.

Given that Glenda receives a salary of <u>$4,792 bi-monthly</u>, we are required to find the earnings per year.

There are <u>12 months in each year</u>, and if the given value is a 2-month value, then we shall divide 12 by 2 to find that there is in total 6 of the 2-month value.

12/2=6

Finally, we shall <u>multiply the total number with the salary</u>, which will be $4,792 times 6.

4792*6=\boxed{28752}

Hope this helps!! :)

Please let me know if you have any questions

3 0
2 years ago
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