Write down all the even numbers from 30 to 300. How many times will the digit "6" appear?
Answers
The correct statement will be that the there are 135 even numbers ranging from 30 to 300 inclusive of the numbers. The digit "6" will be repeated for a number of 63 times.
The calculation of repeated digit "6" can be done by applying simple calculations or even can be calculated on the fingers.
- Even numbers are referred to as those numbers whose last digit ends with either 0,2,4,6 or 8. There are a total of 270 digits between and half of them will be even. So even digits are 135 inclusive of both the numbers.
- The calculation where 6 is repeated is done by applying the information that there will be 17 times 6 appears between values 30 to 99.
- Between 100 to 200 and 201 to 300, "6" is repeated 23 times in each of the ranges.
- Calculation of "6" is as below,
- So we now know that there are 135 digits that are even and 63 times the digit 6 is repeated in the range 30 thru 300.
Hence, the correct statement is that there are 135 even digits between the range of 30 to 300 and digit 6 is repeated for a number of 63 times.
To know more about numerical ranges, click the link below.
Step-by-step explanation:
In the Numbers from 30 to 99 the 6 will be written 1*4 + 6*1 + 1*1*2 = 12
In the Numbers from 100 to 300 the 6 will be written 0 + 3*1*4 + 3*9*1 + 3*1*1*2 = 45
So the six will be written 12 + 45 = 57 times
Is this right? Do you have the same result?
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Answers
The answer is: 8/27 or 0.2963.
Answers
Answer:
x = 9
Step-by-step explanation:
Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Divide 0.7 from both sides:
6.3 = 0.7x
(6.3)/0.7 = (0.7x)/0.7
x = 6.3/0.7
x = 9
x = 9 is your answer.
~
Answer:
9
Step-by-step explanation:
6.3/.7=x
9=x
Answers
First find the area of the whole trapezoid, or W but it already gives us that as 21.66 in squared.
Next, find the area of the non shaded region, or NS.
3.8*4.6=17.48 in
Lastly, subtract the NS from the W.
21.66 -17.48 =
NS= 17.48 in squared.
W= 21.66 in squared.
S= 4.18 in squared.
The answer is B- 4.18 in Squared
Answers
Answer:
The following table list two investment plans, A and B. Given this information, determine which investment is an ordinary annuity and the future value of the ordinary annuity after one year, given that both investments, A and B, compound interest monthly at the rate of 3.5%. Round to the nearest cent.

a.
Investment A is an ordinary annuity with $3,918.03 in the account after 1 year.
b.
Investment B is an ordinary annuity with $3,918.03 in the account after 1 year.
c.
Investment A is an ordinary annuity with $3,906.64 in the account after 1 year.
d.
Investment B is an ordinary annuity with $3,906.64 in the account after 1 year
Step-by-step explanation:
IT IS D
Answer:
Investment B is an ordinary annuity with $3,906.64 in the account after 1 year
Step-by-step explanation:
yards of sod. Give the dimensions of two
different rectangular regions that she can
cover with the sod. What is the perimeter
of each region?
Answers
Answer:
The dimensions of the two different rectangular regions are;
1st Arrangement:
W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards
2nd Arrangement:
W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards
The perimeter of the two different rectangular regions are;
1st Arrangement:
P₁ = 18 yards
2nd Arrangement:
P₂ = 24 yards
Step-by-step explanation:
Bella is putting down patches of sod to start a new lawn.
She has 20 square yards of sod.
We are asked to provide the dimensions of two different rectangular regions that she can cover with the sod.
Recall that a rectangle has an area given by
Area = W*L
Where W is the width of the rectangle and and L is the length of the rectangle.
Since Bella has 20 square yards of sod,
20 = W*L
There are more than two such possible rectangular arrangements.
Out of them, two different possible arrangements are;
1st Arrangement:
20 = (4)*(5) = (5)*(4)
Width is 4 yards and length is 5 yards or width is 5 yards and length is 4 yards
2nd Arrangement:
20 = (2)*(10) = (10)*(2)
Width is 2 yards and length is 10 yards or width is 10 yards and length is 2 yards
Therefore, the dimensions of two different rectangular regions are;
1st Arrangement:
W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards
2nd Arrangement:
W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards
What is the perimeter of each region?
The perimeter of a rectangular shape is given by
P = 2(W + L)
Where W is the width of the rectangle and and L is the length of the rectangle.
The perimeter of the 1st arrangement is
P₁ = 2(4 + 5)
P₁ = 2(9)
P₁ = 18 yards
The perimeter of the 2nd arrangement is
P₂ = 2(2 + 10)
P₂ = 2(12)
P₂ = 24 yards
So the perimeter of the 1st arrangement is 18 yards and the perimeter of the 2nd arrangement is 24 yards.
Note:
Another possible arrangement is,
20 = (1)*(20) = (20)*(1)
Width is 1 yard and length is 20 yards or width is 20 yards and length is 1 yard.