(-1 + 6)^2 <span>+ (4+5)^2 = </span>106
Answer:
66 ≤ f ≤100
Explanation
Mean= ( Σ x ) / n
Mean= sum of scores/ number of subject she took
Now, she already too 3 subject which sum is 85+83+86=254
Now we need to know range of score for her to have (grade) a mark between 80 and 89
Now let take the lower limit mean=80
The lowest score she can get is
Mean = ( Σx) / n
80=(85+83+86+f)/4
80×4= 254+f
Therefore, f= 320-254=66
Therefore the minimum score she can have to have a B is 66.
Then, let take the upper limit mean 89. i.e the maximum she can have so that she don't have an A grade.
Mean = ( Σx) / n
89=( 83+85+86+f)/4
89×4= 254+f
f= 356-254
f=102.
Therefore this shows that she cannot have an A grade in the exam. The maximum score for the exam is 100.
There the range of score is 66 ≤ f ≤100 to have a B grade
66 ≤ f ≤100 answer
Since she cannot score 102 in the examination.
Answer:
The number of
Fiction books = x = 2400 fiction books
Non fiction books = y = non fiction books
Step-by-step explanation:
Let the number of
Fiction books = x
Non fiction books = y
The libary has 4,000 books in total.
Hence: x + y = 4000........ Equation 1
The number of fiction books in a libary is %150 of the number of non-fiction books.
x = 150% of y
x = 150/100 × y
x = 1.5y
We substitute 1.5y for x in Equation 1
x + y = 4000
1.5y + y = 4000
2.5y = 4000
y = 4000/2.5
y = 1600 non fiction books
Solving for x
x = 1.5y
x = 1.5 × 1600
x = 2400 fiction books.
Therefore,
The number of
Fiction books = x = 2400 fiction books
Non fiction books = y = non fiction books
It gave you the input because it is telling you what x is
Answer:
Area of field = 10 cm²
Step-by-step explanation:
Given field is a trapezium;
Parallel sides of given field are; 1¹/₂ cm and 3¹/₂ cm
Height of trapezium = 4 cm
Find:
Area of field
Computation:
Area of field = Area of trapezium
Area of trapezium = [Sum of Parallel sides / 2][h]
Area of field = [(1¹/₂ + 3¹/₂) / 2][4]
Area of field = [(5) / 2][4]
Area of field = [2.5][4]
Area of field = 10 cm²
