The scientist that made a more concentrated salt solution is scientist A.
<h3>How to compute the value?</h3>
Since Scientist A dissolved 1.0 kilogram of salt in 3.0 liters of water. The concentration will be:
= 1/3 = 0.33
Scientist B dissolved 2.0 pounds of salt into 7.0 pints of water. The concentration will be:
= 2/7
= 0.285
Therefore, the scientist that made a more concentrated salt solution is scientist A.
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Pythagorean triples are integer solutions to the Pythagorean theorem a^2 + b^2 = c^2 I like triplets but triples seems to be the favored term for right triangle the x side is the hypotenuse the side opposite the right angle
hope this helps!
Answer:
$397.34 (if he sold the 20 leftover hot dogs), $297.34 if he didn't.
Step-by-step explanation:
We are going to assume that a month has 30 days.
- First, we are going to see how much money the vendor got from selling the 80 hot dogs. He sold 80 hot dogs at 20 dollars/piece = 1600 dollars.
- We need to subtract the amount of money he spent in each hot dog (12 dollars in raw material plus one dollar for packing): 13 dollars x 100 hot dogs he prepared = 1300 dollars
- He also spends a total of 80 dollars per month in truck rent, electricity and other expenses. If we divide this by the amount of days per month we have: 80/30 = 2.66
- The problem doesn't tell us that there were unhappy customers that day so that amount is zero.
- We are going to assume that the vendor sold the remaining 20 hot dogs at 5 dollars/piece. 20 x 5 = 100.
Thus, the profit for that day is:
1600 - 1300 - 2.66 + 100 = 397.34
<u>(</u><u>Note:</u><u> If the vendor did not sell the leftover hot dogs and he actually only sold 80 hot dogs, then the profit would be: 1600 - 1300 - 2.66 = 297.34)</u>
Answer:
Step-by-step explanation:
1/4 x 20x - 1/4 x 28<2x - 6 + 2
5x - 28/4<2x - 4
5x - 2x<-4 + 28/4
3x<-4/1 + 28/4
3x<-16 + 28/4
3x<12/4
3x<3
3x/3<3/3
x<1
Answer: D. minimizes the sum of the squared residuals
Step-by-step explanation: The ordinary least square method is often used in locating the trendine which best fits a graphical linear model. The best is one in which the sum of the squared residual is smallest. The residual refers to the difference between the actual and the predicted points. The sum of the squared differences is obtained and the trend line is positioned where the residual is minimum. Choosing a OLS, and minimizing the sum.of the squared residual, the error difference between the predicted and actual score is minimized or reduced, hence, improving the prediction accuracy of our model.
