22 play violin.
1/5 * 40 = 8
1/4 * 40 = 10
8+10=18
40-18=22
22!
SOLUTION:
Amount added / Original amount × 100% = % increase
14 / 25 × 100 = 56%
FINAL ANSWER:
Therefore, the percentage increase in coins is 56%.
Hope this helps! :)
Translate to english and i’ll help
Answer:
Option C
Step-by-step explanation:
You forgot to attach the expression that models the cost of the camping trip during the three days. However, by analyzing the units, the answer can be reached.
The total cost has to be in units of $.
There are two types of costs in the problem:
Those that depend on the number of days ($/day
)
Those that depend on the number of students and the number of days ($/(student * day))
If there are 3 days of camping and b students, then you have to multiply the costs that depend on the days by the number of days (3), and the costs that depend on the number of students you have to multiply them by 'b'
So, if the costs that must be multiplied by 'b' are only those that depend on the number of students, the coefficient of b must be:
3 days (Cost of training + Cost of food Miscellaneous expenses :).
Therefore the correct answer is option C:
C. It is the total cost of 3 days per student of Mr. Brown, with training, food and miscellaneous expenses.
The expression that represents the total expense should have a formula similar to this:
![y = (3 days) *([\frac{20.dollars}{(day * student)} + \frac{30.dollars}{(student * day)} + \frac{50.dollars}{(student * day)}] b + \frac{200}{day}) + 1050.dollars](https://tex.z-dn.net/?f=y%20%3D%20%283%20days%29%20%2A%28%5B%5Cfrac%7B20.dollars%7D%7B%28day%20%2A%20student%29%7D%20%2B%20%5Cfrac%7B30.dollars%7D%7B%28student%20%2A%20day%29%7D%20%2B%20%5Cfrac%7B50.dollars%7D%7B%28student%20%2A%20day%29%7D%5D%20b%20%2B%20%5Cfrac%7B200%7D%7Bday%7D%29%20%2B%201050.dollars)
y = 3 ($ 100b + $ 200) + $ 1050
Answer:
And the 90% confidence interval would be between [566.80 and 579.20]
Step-by-step explanation:
Information given
represent the sample mean
population mean (variable of interest)
s=22 represent the sample standard deviation
n=36 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
(1)
The degrees of freedom are given by:
The Confidence level is 0.90 or 90%, the significance would be
and
, and the critical value for this case would be
And replacing we got:
And the 90% confidence interval would be between [566.8033 and 579.1967]