Multiply −1.7%0D%0A−%0D%0A1.7%0D%0A by 2 1/2%0D%0A2%0D%0A1%0D%0A2%0D%0A. Write the answer as a mixed number in simplest form.(1 point)%0D%0AResponses%0D%0A%0D%0A−234100%0D%0A−%0D%0A2%0D%0A34%0D%0A100%0D%0Anegative 2 Start Fraction 34 over 100 End Fraction%0D%0A%0D%0A−414%0D%0A−%0D%0A4%0D%0A1%0D%0A4%0D%0Anegative 4 Start Fraction 1 over 4 End Fraction%0D%0A%0D%0A−4520%0D%0A−%0D%0A4%0D%0A5%0D%0A20%0D%0Anegative 4 Start Fraction 5 over 20 End Fraction%0D%0A%0D%0A−174%0D%0A−%0D%0A17%0D%0A4%0D%0Anegative Start Fraction 17 over 4 End Fraction%0D%0ACheck answerRemaining Attem

1 answer

To solve the problem, we need to multiply \(-1.7%\) by \(2 \frac{1}{2}%\).

First, let's convert both percentages to fractions:

  1. Convert \(-1.7%\) to a fraction: \[ -1.7% = -\frac{1.7}{100} = -\frac{17}{1000} \]

  2. Convert \(2 \frac{1}{2}%\) to a fraction: \[ 2 \frac{1}{2}% = \left(2 + \frac{1}{2}\right)% = \frac{5}{2}% \] \[ \frac{5}{2}% = \frac{5}{2} \times \frac{1}{100} = \frac{5}{200} = \frac{1}{40} \]

Now we can multiply the two fractions: \[ \text{Multiplication: } -\frac{17}{1000} \times \frac{1}{40} = -\frac{17 \times 1}{1000 \times 40} \] Calculating the denominator: \[ 1000 \times 40 = 40000 \]

Then, the multiplication becomes: \[ -\frac{17}{40000} \]

Next, we will write \(-\frac{17}{40000}\) as a mixed number. Since the numerator \(17\) is less than \(40000\), the whole number part is \(0\), and we will focus on the fractional part: \[ -\frac{17}{40000} \]

Now, we need to simplify \(\frac{17}{40000}\):

  • \(17\) is a prime number, and \(40000\) can be factored into \(2^8 \times 5^4\), so there are no common factors to simplify further.

Thus, the simplest form of the fraction is \(-\frac{17}{40000}\).

Since there is no whole number part, the answer as a mixed number is: \[ 0 - \frac{17}{40000} = -\frac{17}{40000} \]

The answer as a mixed number in simplest form is: \[ -\frac{17}{40000} \]

(Note: No available option in your list matches \(-\frac{17}{40000}\), so verify the available options, as they seem incorrect.)