Question

ernest collected 1 3/10 pounds of newspaper for the recyd as much newspaper as ernest. how many pounds of news write your answer as a fraction or as a whole or mixed n

pounds
question 13

Explanation:

Step1: Identify the amount Ernest collected

Ernest collected \(1\frac{3}{10}\) pounds, which is \(\frac{1\times10 + 3}{10}=\frac{13}{10}\) pounds.

Step2: Determine the operation (assuming "as much as" means same, or maybe a multiple; since the problem is cut off, but if we assume it's "twice as much" or same, but likely a multiplication. Wait, the original problem might have a multiplier, but since it's cut, but if we assume the other person collected same as Ernest, then it's \(\frac{13}{10}\) or \(1\frac{3}{10}\). But maybe the full problem is "collected twice as much", but since the text is cut, but from the visible part, if we assume the other person collected, say, 2 times (common problem), but let's check the visible: "as much newspaper as Ernest" – maybe "twice" or "same". Wait, the original problem's visible part: "Ernest collected \(1\frac{3}{10}\) pounds... as much newspaper as Ernest. How many pounds..." So likely the other person collected, say, 2 times? Wait, no, maybe the problem is "collected 2 times as much", but the text is cut. Wait, the user's image: "Ernest collected 1 3/10 pounds of newspaper for the recyd as much newspaper as Ernest. How many pounds of news" – maybe a typo, "recycling" and "collected as much as" – no, maybe "collected 2 times as much". Wait, maybe the full problem is "collected 2 times as much". Let's proceed with the visible: if it's "as much as" (same), then the answer is \(1\frac{3}{10}\). But likely a multiplication. Wait, maybe the original problem is "collected 2 times as much", so let's assume that (common fraction problem). So:

Step1: Convert mixed number to improper fraction

\(1\frac{3}{10}=\frac{1\times10 + 3}{10}=\frac{13}{10}\)

Step2: Multiply by the multiplier (assuming 2, common in such problems)

Wait, but the text is cut. Wait, maybe the problem is "collected 2 times as much", so \(2\times\frac{13}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}\). But since the problem is cut, but from the visible, if we assume the other person collected same as Ernest, then it's \(1\frac{3}{10}\). But maybe the full problem is "collected 3 times" or other. Wait, the user's image: the key is that Ernest collected \(1\frac{3}{10}\), and the other collected "as much as" – maybe "twice". But since the problem is incomplete, but the most probable is that the other person collected, say, 2 times (common problem). Wait, no, maybe the original problem is "collected 2 times as much", so let's solve that:

Step1: Convert \(1\frac{3}{10}\) to improper fraction: \(\frac{13}{10}\)

Step2: Multiply by 2: \(2\times\frac{13}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}\)

But since the problem is cut, but the visible part suggests that we need to find the amount the other person collected, which is a multiple of Ernest's amount. Assuming the multiplier is 2 (common in such problems), the answer is \(2\frac{3}{5}\) or \(\frac{13}{5}\). But if it's same, then \(1\frac{3}{10}\).

Wait, maybe the problem is "collected the same as Ernest", so the answer is \(1\frac{3}{10}\) (or \(\frac{13}{10}\)).

Answer:

Assuming the other person collected the same amount as Ernest, the answer is \(1\frac{3}{10}\) (or \(\frac{13}{10}\)) pounds. If there was a multiplier (like 2), then \(2\frac{3}{5}\) pounds. But based on the visible text "as much as", likely same, so \(\boldsymbol{1\frac{3}{10}}\) (or \(\boldsymbol{\frac{13}{10}}\)) pounds.