Let T represent the fifth test score. The average is the sum of scores divided by their number.
... 82 = (83 +75 +93 +67 +T)/5
... 410 = 318 +T . . . . . . . multiply by 5
... 92 = T . . . . . . . . . . . . subtract 318
The fifth score was 92.
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One of the ways I like to work problems like these is to add the differences from the mean. Those are ...
1, -7, 11, -15
so their sum is -10.
The final score must be 10 above the mean in order for this total to be zero. 10 above the mean is 92.
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Step-by-step explanation:
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its upsidedown what are you doing
Distance ran by Rudy on Monday = 2 km
Percentage of increase in distance of running each day = 15%
Let us assume that the day on which Rudy will run 4 km = x
Then
2 + 0.15x = 4
0.15x = 2
x = 2/0.15
= 13.3
From the above deduction we can conclude that Rudy will run more than 4 km on the 14th day or on Sunday. I hope the procedure is clear enough for you to understand.
Answer:
6.784 × 10¹⁰
Step-by-step explanation:
→ First count the number of 0's
0000000 = 7 zero's
→ Count how many decimal places until a single digit
6.784 = 3 decimal places
→ Add the number of zero's with the decimal place's
7 + 3 = 10
6.784 × 10¹⁰
