Answer:
542 minutes
Step-by-step explanation:
currently our mean is =530
i got that by adding all the number out which equal too =2,650
and divide it by how many number there are which is 5
2,650/5=530
After some trial and error i found it
so you add 494+690+502+478+486+542=3,192
3,192/6=532
and that was our goal
so the 6th month she talked on the phone for 542 minutes
pls mark me brainliest
Answer:
You are more likely to win by playing regular defense.
Step-by-step explanation:
Assume out of 100 reviewed games, there were 50 regular defense games and 50 prevent defense games. And out of 50 regular defense games, 38 were win, 12 were lose. And out of 50 prevent defense game, 29 were win, 21 were lose.
Probability to win the game by playing regular defense is:
P(win | regular) = 38/50 = 0.76
Probability to win the game by playing prevent defense is:
P(win | prevent) = 29/50 = 0.58
Since the probability of winning by regular defense game is more than prevent defense game (0.76 > 0.58), you are more likely to win by playing regular defense.
Hello :) ok so to find AC you would have to use sin. It would be sin70= x/4. To find AC you would have to multiply 4 by sin70 which is 3.758 or to the nearest hundredth would be 3.76. I hope I helped, if you need a more in depth explanation lmk
Step-by-step explanation:
sorry i couldn't solve the last and the last second one
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What is it?
The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
How do you find IQR?
<em>Step 1: Put the numbers in order. ...</em>
<em>Step 2: Find the median. ...</em>
<em>Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. ...</em>
<em>Step 4: Find Q1 and Q3. ...</em>
<em>Step 5: Subtract Q1 from Q3 to find the interquartile range.</em>
