The original price of the hat is $35. Paolo paid $28 for the hat.
So discount for the hat is $(35-28) = $7
$7 is discounted from $35. We have to find the percentage of the discount.
To find the percentage we have to divide the discount amount by the original price and then have to multiply it by 100.
So the discount percentage =
((7/35)×100) %
We can simplify 7/35 by dividing 7 and 35 both by 7. So we will get 1/5 after simplifying.
((1/5) ×100) %
(100/5)% = 20%
So the hat was discounted by 20%.
Answer:
-1 17/25
Solution with Steps
−65−1225=?
The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.
LCD(-6/5, 12/25) = 25
Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.
(−65×55)−(1225×11)=?
Complete the multiplication and the equation becomes
−3025−1225=?
The two fractions now have like denominators so you can subtract the numerators.
Then:
−30−1225=−4225
This fraction cannot be reduced.
The fraction
−4225
is the same as
−42÷25
Convert to a mixed number using
long division for -42 ÷ 25 = -1R17, so
−4225=−11725
Therefore:
−65−1225=−11725
Answer: A
Step-by-step explanation: $88.74÷9=$9.86
$88.74 is the total amount
9 players, and each player has to pay the same amount
we have to divide it evenly in 9 pieces, so we divide $88.74 into 9 groups which then equals $9.86
Hope this helps!
Answer:
66
Step-by-step explanation:
In statistics, the formula for RANGE is given as the difference between the Highest and the Lowest value.
In the above values we are given data consisting of the 7 games that Roger bowled.
155, 165, 138, 172, 127, 193 , 142.
Step 1
We arrange from the least to the highest.
127, 138, 142, 155, 165, 172, 193
Step 2
Lowest value = 127
Highest value = 193
Step 3
Range = 193 - 127
= 66
Therefore, the range of Roger's scores is 66
The area of a square is expressed as the length of the side to the power of two, A = s^2. We were given the area of the enlarged photo which is 256 in^2. Also, it was stated that the length of the enlarged photo is the length of the original photo plus ten inches. So, from these statements we can make an equation to solve for x which represents the length of the original photo.
A = s^2
where s = (x+10)
A = (x+10)^2 = 256
Solving for x,
x= 6 in.
The dimensions of the original photo is 6 x 6.
