There are 45 new houses being built in a neighborhood. Last month, 2/3 of them were sold. This month 2/5, of the remaining houses were sold. How many houses are left to be sold?
Answers
Answer: 9 houses
Step-by-step explanation:
Step 1:
Since 2/3 of the houses were sold last month, 1/3 of them were left.
45 * 1/3 = 15 houses left last month
Step 2:
Since 2/5 of the remaining houses were sold this month, 3/5 of them were left
15 * 3/5 = 9 houses left to be sold
there's your answer hope this helps :)
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Please solve it from the picture below!!
Answers
Good luck!!!
select the explicit function which defines the sequence.
A.) f(n) = 1/3 • 2^(n - 1)
B.) f(n) = 2 • (1/3)^(n - 1)
C.) f(n) = 1/3 • 3^(n - 1)
D.) f(n) = 3 • (1/3)^(n - 1)
Answers
The explicit formula that defines the sequence is
EXPLANATION
The given sequence is
The first term of this sequence is
We can find the common ratio by expressing a subsequent term over a previous term and simplifying it.
The common ratio is
The formula for finding the nth term of the given geometric sequence is given by,
We now substitute the value of the first term and the common ratio in to the above formula to obtain,
The correct answer is option D.
Answer:
D
Step-by-step explanation:
Answers
The average age of the employees in 2003 is 57.216 years. And, the average age of the employees in 2009 is 59.184 years.
Given that;
The function A(s) given by ,
A (s) = 0.328s + 50
Now for the average age of employees in 2003 and 2009 using the function A(s) = 0.328s + 50, substitute the values of s into the equation.
For the year 2003,
Since s represents the number of years since 1981,
Hence, subtract 1981 from 2003:
s = 2003 - 1981
s = 22
Now substitute this value of s into the function A(s):
A(22) = 0.328 × 22 + 50
A(22) = 7.216 + 50
A(22) = 57.216
Therefore, the average age of the employees in 2003 is 57.216 years.
Similarly, for the year 2009,
s = 2009 - 1981
s = 28
Substituting this value into the function:
A(28) = 0.328 × 28 + 50
A(28) = 9.184 + 50
A(28) = 59.184
Hence, the average age of the employees in 2009 is 59.184 years.
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Final answer:
The mathematical problem involves calculating the average age of employees at a company for the years 2003 and 2009 using the linear function A(s), where 'A(s)' represents the average age and 's' is the number of years since 1981. The calculated average ages for the employees in the years 2003 and 2009 are approximately 57 and 59 years, respectively.
Explanation:
The subject is mathematics, specifically linear functions. Based on the equation A(s) = 0.328s + 50, where 'A(s)' represents the average age of the employees and 's' represents the number of years since 1981. In the year 2003, s would be 22 (2003-1981) and in 2009, s would be 28 (2009-1981).
Substituting these values of 's' into the function gives:
For 2003, A(22) = 0.328*22 + 50 = 57.216
For 2009, A(28) = 0.328*28 + 50 = 59.184
Therefore, the average age of the employees at the company in 2003 and 2009 were approximately 57 and 59 years, respectively.
Learn more about Linear Functions here:
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Answers
Answer:
70
Step-by-step explanation:
Answer:
70
Step-by-step explanation:
3(2)+2=8
3(3)+2=11
3(4)+2=14
3(5)+2=17
3(6)+2=20
Sum=70
Answers
Answer:
The answer is
Step-by-step explanation:
we know that
A quotient is the division of two numbers
In the expression divided by n
the numerator is equal to
the denominator is equal to
so
the quotient is equal to
The complete expression divided by n plus
is equal to
Hope that helps
Answers
x - the shortest side
2x
x + 28
The sum of these must be equal to the perimeter of the flower bed, so
x + 2x + x + 28 = 184
4 x + 28 = 184 Combined like terms
4x = 156 Subtracted 28 from both sides
x = 39 Divided both sides by 4.
So the dimensions are 39 feet, 78 feet, and 67 feet