1/4 x 4/7
do the top numbers first. 1 x 4 is 4do bottom numbers next 4 x 7 is 28
that leaves you with 4/28. But you have to reduce.
4/28 = 2/14 = 1/7 (notice I just cut both the top and bottom numbers in half each time I reduced until I couldn't reduce any more).
So, final answer is 1/7
Just divide all the numbers and u should get your answer.
If 101 is given in base <em>b</em>, then in base 10 you have
For instance, if <em>b</em> = 2, then
101₂ = 2² + 1 = 5₁₀
If <em>b</em> = 3, then
101₃ = 3² + 1 = 9₁₀
If <em>b</em> = 4, then
101₄ = 4² + 1 = 17₁₀
The answer depends entirely on the base of the given number.
Answer:
C) About $85
Step-by-step explanation:
Please consider the complete question.
Carl puts $1.10 in his penny bank every day in the month of July (31 days). His total saving at the end of June is $55.00. What is the best estimate at the end of July?
A) about $25
B) about $115
C) about $85
D) about $66
First of all, we will round $1.10 to nearest whole number and round 31 to nearest tens as:
So amount saved in 30 days would be 30 times $1.
Carl's savings at the end of July would be amount saved in June plus amount saved in July that is 
Therefore, the estimate for Carl's saving at the end of July is $85 and option C is the correct choice.
Step-by-step explanation:
what is to be done ? just that addition ?
first, let's all bring to pure fractions.
7 2/3 = (7×3 + 2)/3 = 23/3
2 1/5 = (2×5 + 1)/5 = 11/5
then let's bring all to a common denominator (bottom part of the fraction).
for this, we need to find the first (or any) common multiple of the 2 denominators (here 3 and 5).
if you have no other idea, just multiply them with each other (in our case here this is also the smallest common multiple) : 3×5 = 15.
so, we bring both fractions into the form of .../15. then it is easy to add them.
23/3 = x/15
x = 23/3 × 15 = 23×15/3 = 23×5 = 115
=>
23/3 = 115/15
11/5 = x/15
x = 11/5 × 15 = 11×15/5 = 11×3 = 33
=>
11/5 = 33/15
now the addition :
115/15 + 33/15 = 148/15 or 9 13/15
this is the result, as it cannot be further simplified.
