Answer:
There were 333 seniors in 1990
153 seniors had jobs in 2010.
Step-by-step explanation:
Let the number of high school seniors in 1990 = x
75% of x had jobs in 1990
In 2010 there were 50 more seniors. That is, the number of seniors is x+50.
40% of (x+50) had jobs in 2010.
Total number of jobs in both years = 403.
\[0.75 * x + 0.4 * (x + 50) = 403\]
\[0.75 * x + 0.4 * x + 20 = 403\]
=> \[1.15 * x = 403 - 20\]
=> \[1.15 * x = 383\]
=> \[x = 383/1.15\]
=> \[x = 333\]
Number of seniors having job in 2010 = 0.4 * (333 + 50) = 0.4 * 383 = 153.2 = 153(approx)
Answer: First of all, we will add the options.
A. Yes, because 3 inches falls above the maximum value of lengths in the sample.
B. Yes, because the regression equation is based on a random sample.
C. Yes, because the association between length and weight is positive.
D. No, because 3 inches falls above the maximum value of lengths in the sample.
E. No, because there may not be any 3-inch fish of this species in the pond.
The correct option is D.
Step-by-step explanation: It would not be appropriate to use the model to predict the weight of species that is 3 inches long because 3 inches falls above the maximum value of lengths in the sample.
As we can see from the question, the model only accounts for species that are within the range of 0.75 to 1.35 inches in length, and species smaller or larger than that length have not been taken into consideration. Therefore the model can not be used to predict the weights of fishes not with the range accounted for.
<em>Answer:</em>
25.4639073013
Rounded is 25.4
<em>Have a beautiful day! </em>v(⌒o⌒)v♪
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I dont undertand the go negative for the x and y but i can't read that so pick the -x,-y or something
15 divide 30 by 2 and you have your answer
