D=dimes
q=quarters
if she has 12 coins then
12-d=q she has 12-d quarters
we dont know how much money she has unless we know d.
she could have 12 dimes and 0 quarters or 11 dimes and 1 quarter etc. we need more information to solve.
Okay I think there has been a transcription issue here because it appears to me there are two answers. However I can spot where some brackets might be missing, bear with me on that.
A direct variation, a phrase I haven't heard before, sounds a lot like a direct proportion, something I am familiar with. A direct proportion satisfies two criteria:
The gradient of the function is constant s the independent variable (x) varies
The graph passes through the origin. That is to say when x = 0, y = 0.
Looking at these graphs, two can immediately be ruled out. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations.
This leaves B and C. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear.
This leaves C as the odd one out.
I hope this helps you :)
20/28 = 3x / (4x + 2)
84x = 20(4x + 2)
84x - 80x = 40
4x = 40
x = 10
so the 2 lines are equal to 30 and 42
Answer:
Step-by-step explanation:
Let the total number of Newspapers be x
Number of newspapers delivered in first hour of his route= 
of x
Total number of newspapers delivered in first hour= x
Number of newspapers left= x-
Number of newspapers left=
Number of newspapers left=
Number of newspapers delivered in second hour=4/5 of
Number of newspapers delivered in second hour=
Fraction of newspapers delivered in second hour is=
Hence, the correct answer is
Answer:
Since the null hypothesis is true, finding the significance is a type I error.
The probability of the year I error = level of significance = 0.05.
so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5
Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.
