The LCD is 1/15 and 3/15.
Answer:
where is the statement?????!!!
The value of x that completes the conditional relative frequency table by column is 0.85
<h3>How to determine the value of x?</h3>
From the complete question, we have the following column elements
0.15 x 1
Express a sum
0.15 + x = 1
Subtract 0.15 from both sides
x = 0.85
Hence, the value of x that completes the conditional relative frequency table by column is 0.85
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The restrictions on the variable of the given rational fraction is y ≠ 0.
<h3>The types of numbers.</h3>
In Mathematics, there are six (6) common types of numbers and these include the following:
- <u>Natural (counting) numbers:</u> these include 1, 2, 3, 4, 5, 6, .....114, ....560.
- <u>Whole numbers:</u> these comprises all natural numbers and 0.
- <u>Integers:</u> these are whole numbers that may either be positive, negative, or zero such as ....-560, ...... -114, ..... -4, -3, -2, -1, 0, 1, 2, 3, 4, .....114, ....560.
- <u>Irrational numbers:</u> these comprises non-terminating or non-repeating decimals.
- <u>Real numbers:</u> these comprises both rational numbers and irrational numbers.
- <u>Rational numbers:</u> these comprises fractions, integers, and terminating (repeating) decimals such as ....-560, ...... -114, ..... -4, -3, -2, -1, -1/2, 0, 1, 1/2, 2, 3, 4, .....114, ....560.
This ultimately implies that, a rational fraction simply comprises a real number and it can be defined as a quotient which consist of two integers x and y.
<h3>What are
restrictions?</h3>
In Mathematics, restrictions can be defined as all the real numbers that are not part of the domain because they produces a value of 0 in the denominator of a rational fraction.
In order to determine the restrictions for this rational fraction, we would equate the denominator to 0 and then solve:
23/7y;
7y = 0
y = 0/7
y ≠ 0.
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Complete Question:
State any restrictions on the variables 23/7y
The average decrease in value can be solve by first solving the value at year 1 and year 2
and i think the equation is <span>f(x)=10,000(0.73)^x</span>
at x = 1
<span>f(x) = 10,000(0.73)^x
</span><span>f(1) = 10,000(0.73)^(1)
f(1) = 7300
at x = 2
</span><span>f(x) = 10,000(0.73)^x
</span><span>f(2) = 10,000(0.73)^(2)
f(2) = 5329
so the average decrease = ( 7300 - 5329) = $ 1971 per year</span><span>
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