PLEASE HELP WILL MARK BRAINLIEST Which is the solution to the inequality?

2 and three-fifths less-than b minus StartFraction 8 over 15 EndFraction

A. b less-than 2 and StartFraction 1 over 15 EndFraction

B. b greater-than 2 and StartFraction 1 over 15 EndFraction

C. b less-than 3 and StartFraction 2 over 15 EndFraction

D. b greater-than 3 and StartFraction 2 over 15 EndFraction

Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

or

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Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality is the same as

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Let's multiply both sides by 15 to clear out the fractions

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Now isolate the variable b

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

i'm assuming it's looking for the missing length, which should be 4cm

hope this !

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.