To solve this question, simply use the formula to find the volume of the cylinder,
V = Pi r^2 • h
V = 3.14 • 4^2 • 4.
Then compare the value for volume to the closest number on the number line. Is it in between 2 numbers, exactly one number on the number line or very close to one number as in the number line.
Answer:
Step-by-step explanation:
Assuming there is a punitive removal of one point for an incorrect response.
Five undiscernable choices: 20% chance of guessing correctly -- Expectation: 0.20*(1) + 0.80*(-1) = -0.60
Four undiscernable choices: 25% chance of guessing correctly -- Expectation: 0.25*(1) + 0.75*(-1) = -0.50
I'll use 0.33 as an approzimation for 1/3
Three undiscernable choices: 33% chance of guessing correctly -- Expectation: 0.33*(1) + 0.67*(-1) = -0.33 <== The approximation is a little ugly.
Two undiscernable choices: 50% chance of guessing correctly -- Expectation: 0.50*(1) + 0.50*(-1) = 0.00
And thus we see that only if you can remove three is guessing neutral. There is no time when guessing is advantageous.
One Correct Answer: 100% chance of guessing correctly -- Expectation: 1.00*(1) + 0.00*(-1) = 1.00
Answer:
3.9 mi/h
Step-by-step explanation:
If the boy is rowing perpendicular to the current, the two vectors form a right triangle.
AB represents the downstream current, BC is the speed across the river, and AC is the ground speed of the boat
AC^2 = 2.4^2 + 3.1^2 =5.76 + 9.61 = 15.37
AC = sqrt(15.37) = 3.9 mi/h
The boat's speed over the ground is 3.9 mi/h.
Answer:
11/3
<em>Alternative Form: </em>3 2/3, 3.6
Step-by-step explanation:
44/12
Dive the numerator and denominator by 4
44/4 / 12/4
11/ 12/4
11/3
Step-by-step explanation:
1