The cost of the shirts is stable, so we can subtract $1150 - $84, which = $1066. Now we want to find how many watches we can buy with $1066, or how many times $99 fits into $1066. So divide 1066 by 99 to get 10.77. You can't buy 77ths of a watch, so round it down to 10 and you have your answer. Samantha can buy 10 watches.
Answer:
The answer of the following question is m = \frac{C - b - bt}{r + rt}.
Solution:
C = (b + rm)(1 + t),
C = b + rm + bt + rmt
C = b + bt + rm + rmt
C - b - bt = m (r + rt)
\frac{C - b - bt}{r + rt} = m
t\neq -1,
r\neq 0
Answer:
12-5.5=h
12-5.5=6.5
Kyle needs to volunteer 6.5 more hours.
Step-by-step explanation:
Analysis to obtain the function that models the polulaiton ob bees:
1) First year 9,000 bees
2) Second year: decrease 5% => 9,000 - 0.05* 9,000 = 9,000 * (1 - 0.05) = 9,000 * 0.95
3) Every year the population decreases 5% => 9,000 * 0.95)^ (number of years)
4) if you call x the number of years, and f(x) the function that represents the number of bees, then: f(x) = 9,000 (0.95)^ x.
Analysis of the statements:
<span>1) The
function f(x) = 9,000(1.05)x represents the situation.
FALSE: WE DETERMINED IT IS f(x) = 9,000 (0.95)^x
2) The function
f(x) = 9,000(0.95)x represents the situation.
TRUE: THAT IS WHAT WE OBTAINED AS CONCLUSION OF THE PREVIOUS ANALYSIS.
3) After 2 years, the farmer
can estimate that there will be about 8,120 bees remaining.
Do the math:
f(2) = 9,000 * (0.95)^2 = 9,000 * 0,9025 = 8,122
So, the statement is TRUE
4) After 4
years, the farmer can estimate that there will be about 1,800 bees
remaining.
f(4) = 9,000 * (0.95)^4 = 9,000 * 0.81450625 = 7,330
So, the statement is FALSE
5) The domain values, in the context of the situation, are
limited to whole numbers.
FALSE: THE DOMAIN VALUES ARE ALL NON NEGATIVE REAL VALUES. FOR EXAMPLE THE FUNCTION IS WELL DEFINED FOR X = 5 AND HALF
6) The range values, in the context of the
situation, are limited to whole numbers.
TRUE: THERE CANNOT BE FRACTIONS OF BEES
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