Find mean, median and mode of 140, 130, 90, 80, 60, 50, 30, 20, 10, 0
Alecsey [184]
For the mean you would have to add up all the numbers up and divide by the amount of numbers. For your problem you would get 610/10= 61.
For the median you would arrange all the numbers in order from least to greatest and cancel one from the left and one from the right each time. Since there is no middle number, the closest would be 50 and 60. Thus you you add 50+60 and get 110 and then divide by 2 giving you 55.
For the mode, there is no mode because none of the numbers repeat more than once.
I hope this clarifies your doubt. :)
Answer:
Step-by-step explanation:
It would be 5/6 because the denominator Is smaller than 10
can I be brainliest
Answer:
20.25
Step-by-step explanation:
Percentage solution with steps:
Step 1: Our output value is 135.
Step 2: We represent the unknown value with $x$
.
Step 3: From step 1 above,$135=100\%$
.
Step 4: Similarly, $x=15.\%$
.
Step 5: This results in a pair of simple equations:
$135=100\%(1)$.
$x=15.\%(2)$
.
Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both
equations have the same unit (%); we have
$\frac{135}{x}=\frac{100\%}{15.\%}$
Step 7: Again, the reciprocal of both sides gives
$\frac{x}{135}=\frac{15.}{100}$
$\Rightarrow x=20.25$Therefore, $15.\%$ of $135$ is
sorry if it took to long have a great day and brainliest is appreciated!!!!!
Answer:
The cafeteria provides three meals per day.
<u>Reason 1</u>
Yes, they vary directly
As number of days increases ,Total number of meals i.e
1st day ⇒ 3
2nd day⇒6
3 rd day⇒9
4th day⇒12
......................
.........................
increases.
Total number of Meals = k×Number of days
But there is another possibility also
<u>Reason 2</u>
1 st day ⇒ 3
2nd day ⇒3
3rd day⇒ 3
.....................
.....................
As you can see from the above expression On each day number of meals is
constant.
So we can say that ,
On each Day=Constant amount of meal=3
So, there is no Proportionality between Days and Meal.
