Answer:
1,145,375 cm³
Step-by-step explanation:
Like with an earlier question you had, there is a chunk missing. If that chunk was filled in, this would be a 205 * 70 * 85 rectangular prism.
205 * 70 * 85
17,425 * 70
1,219,750
The cut out chunk is a 25 * 70 * 85 triangular prism.
1/2 (25 * 70 * 85)
1/2(2,125 * 70)
1/2(148,750)
74,375
Now that we know what the cut-out chunk is, subtract that from the first value we got.
1,219,750 -74,375
1,145,375 cm³
The volume of the Canadian Post mailbox is 1,145,375 cm³.
Answer:
40
60
Step-by-step explanation:
20÷2=10
10×4=40
10×6=60
Answer:
<h2 /><h2>The interquartile range (IQR) is the difference between the upper (Q3) and lower (Q1) quartiles, and describes the middle 50% of values when ordered from lowest to highest. The IQR is often seen as a better measure of spread than the range as it is not affected by outliers. Interquartile Range. 25% of values.</h2>
<h2 />
<h2>here your answer </h2>
Answer:
1/2 or 50% probability
Step-by-step explanation:
There are just two players, player A and player B. Since it is between this two players, for player A to score first we would have:
AB where player A is the first player to score, we wouldn't have BA since player would not be the first here
Probability = number of favorable outcomes/total number of outcomes
Therefore probability of A being the first player = number of A/total number of players
=1/2 = 0.50
There is therefore a 50% chance of A being the first player
Answer:
0.643
Step-by-step explanation:
Let, s = play a sport
c = participate in a club
P(s) = 0.14
P(s n c) = 0.09
probability that a student participates in
a club given that they also play a sport = P(c | s)
P(c | s) = P(c n s) / P(s)
P(c | s) = 0.09 / 0.14
P(c | s) = 0.64285
= 0.643
