Answer:
4
Step-by-step explanation:
Because i divided 4 2/3 by 22 1/6 = 4.75 and for full sheets it would be 4
Hope This Helped
The math club must sell 50 pies in order to reach the goal of 200 dollars
Deposited amount = $14.
Rate of interest = 2.7%
In decimals 2.7 could be written by dividing 2.7 by 100, we get
0.027.
Interest earned each year = 2.7% of $14 = 0.027*14 = $0.378.
We can make a table to balance by after each year.
First year balance: First year interest + Deposited amount = 0.378 +14 = $14.378.
Second year balance: Second year interest + Deposited amount = 2*0.378 +14 = 0.756+ 14 = $14.756.
Third year balance: Third year interest + Deposited amount = 3*0.378 +14 = 1.134+ 14 = $15.134.
We get four coordinates to graph A(0,14), B(1, 14.378) , C(2, 14.756) and D(3, 15.134).
Plotting those coordinates of the points A, B, C and D on the graph.
Answer:
B. The lower extreme increased.
Step-by-step explanation:
From the original box and whisker plot, the lower extreme (minimum) is 5 ; The upper extreme, maximum is 15.
The number of batches baked on the eight day is 20. This exceeds tbe previous maximum value. Hence, upper extreme( maximum) value of the new plot will change from 15 to 20.
The lower extreme isn't affected as the obtained value isn't below 5.
The median value will Increase and the upper quartile value will also increase. Once the upper quartile value increases, the interquartile range will also increase
Answer: The required solution is 
Step-by-step explanation:
We are given to solve the following differential equation :

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have

Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are

and

Thus, the required solution is 