Answer:
3/5 has the smallest denominator
Step-by-step explanation:
Question:
There exist infinitely many common fractions a/b , where a > 0 and b > 0 and for which 3/5 < a/b< 2/3. Of these common fractions, which has the smallest denominator? Express your answer as a common fraction.
Solution
A Common fraction is a rational number written in the form: a/b. Where a and b are both integers.
The denominator and numerator in this case are greater than zero. That is, they are non zeros.
The least common denominator (LCD) of two non- zero denominators is the smallest whole number that is divisible by each of the denominators.
To find the smallest denominator between 3/5 and 2/3, we would convert the fractions to equivalent fractions with a common denominator by finding their LCM (lowest common multiple).
When comparing two fractions with like denominators, the larger fraction is the one with the greater numerator and the smaller fraction is one with the smaller numerator.
In our solution after comparing, the smaller fraction would have the smallest denominator.
Find attached the solution.
Answer:
end of third year = 1,200*(1.02)^3 = 1,273.45 plus
end of second year = 1,200*(1.02)^2 = 1,248.48 plus
end of first year = 1,200*(1.02) = 1,224.00
Total = 3,745.9
Step-by-step explanation:
Answer:
Hope it helps u
Step-by-step explanation:
As we know that ,
Mean = sum of the terms/ numbers of terms
But here grouped data is given so , we use the formula
Mean=∑[f. m]/ ∑f
where f is frequency and m is mid point of each height ,
Now first we have to find the mid point of each interval, where
midpoint of each interval = (lower boundary + upper boundary)/2
m1=(150+154)/2 = 152
m2=(155+159)/2= 157,now found other by same formula, for each interval
m3= 162
m4= 167
m5=172 Now we find the midpoint of each interval ,so now
∑[f. m]=f1*m1+f2*m2+f3*m3+f4*m4+f5*m5
now putting the values of each frequency for given interval and midpoint of each interval we will get,
∑[f. m]=456+942+1296+167*x+344 = 167*x+3038
Now find,
∑f=f1+f2+f3+f4+f5
∑f=19+x
Now we have,
∑[f. m]=167*x+3038
∑f=19+x
also given mean height=161.6 cm
putt these values in above equation we get,
161.6=
now solve this ,
161.6(19+x)=167*x+3038
3070.4+161.6*x=167*x+3038
3070.4-3038=167*x-161.6*x
32.4=5.4*x
x=32.4/5.4
<h2>
x=6 Ans........</h2>
The coefficient, 0.25, of the expression represents the late fee for every day the book was overdue.
And the constant, 2, represents the fixed rental fee.
The variable, x, represents the number of days Joseph had the book for.
Hope this helps :)
Answer:
Step-by-step explanation:
We have that
and
This implies that;
and
Now find and expresion for
Which is
