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Compare 32/35 and 9/10.
Comparing 32/35 and 9/10 we have 32/35 is greater than 9/10, that is 32/35 > 9/10.
Fraction is the ratio of two numbers. The upper number is called Numerator and the Lower number is called the Denominator.
We know that if the denominators are the same for two fractions then which has the greatest numerator is a greater fraction than the other.
Given the fractions are 32/35, 9/10
To compare this two fractions we have to make denominators equal first.
LCM of 10,35 = 70
Calculating the fractions,
32/35 = (32*2)/(35*2) = 64/70
9/10 = (9*7)/(10*7) = 63/70
Since 64 > 63
So 64/70 > 63/70
Therefore, 32/35 > 9/10
Hence fraction 32/35 is greater than the other fraction 9/10.
Learn more about Fraction here -
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Answer:
She have to pay the bank at the end of the 6 years = 222000$
Step-by-step explanation:
Formula applied in this case where price, interest rate and time duration is mentioned.
Interest = prt/100
Substituting all the given values in the formula.
interest = 15000 * 6 * 8/100
150 * 6 * 8 = 7200
Interest + 15000 = 15000 + 7200 = 222000$ has to pay the total.
In this question amount is 15000 multiply by 6 and by 8 then divide it by 100 so we get the interest then we add it in to amount so we get an answer that is total amount which he has to pay. Hope this helps! :P PLZ GIVE ME BRAINLIST
answer is 2+ 1/4x + 1 the simplest form is x= -3/4 or -0.75
You have to do subtract 10 from 4 because 6 is how much we will profit from each shirt. Now, 6 times 327 is $1,962. That's how much Timothy earned of profit.
Answer:
Option A
Step-by-step explanation:
Given that A linear model is given for the data in the table: y=1.25x+2.
Let us write observed values for each x and also the predicted values as per equation.
x 2 3 4 8 10 16 20 24 Total
y((O) 3 4 7 12 16 22 28 30
y(P) 4.5 5.75 7 12 14.5 22 27 32
DEv 1.5 1.75 0 0 1.5 0 1 2 7 75
where y(0) represents observed y or y in the table given
y(P) gives values of y predicted as per the equation 1.25x+2
Dev represents the absolute difference
Hence answer is option
A.7.75
