The equation to find percent error is:
percent error=(experimental-theoretical)/theoretical *100
So, (115-82.5)/82.5=0.393939*100
39.3939%, or 39.94%
Hope this helps!!
We can used the Simpson's Rule says to approximate the area under a given curve using the following formula:
<span>(Δx/3)[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + ... + 2f(xn-2) + 4f(xn-1) + f(xn)] </span>
<span>The pool is divided into 8 subintervals. We integrate the given function from 0 to 24, while the graph provides values of f(x) at 7 different points. The first value given, 6.2, is NOT f(0). It is f(3). Using Simpson's Rule, and dividing the lake of 24 meters into 8 subintervals, we write the equation: </span>
<span>area = (3/3)[f(0) + 4f(3) + 2f(6) + 4f(9) +2f(12) + 4f(15) + 2f(18) + 4f(21) + f(24)] </span>
<span>Pool area = 0 + 4(6.2) + 2(7.2) +4(6.8) + 2(5.6) + 4(5.0) +2(4.8) +4(4.8) + 0 = 126.4 m^2 </span>
<span>Rounding to the nearest square meter, the area of the lake is approximately 126 m^2 </span>
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.
Answer and Step-by-step explanation:
According to the given situation, The r-value associated with the ordered pairs for the linear function is very nearest to zero which does not results in adequate presentation of the outcome.
A quadratic model could properly comprise of a combination of data, as the set of data has a turning point.
This result in data rises and falls which represents the graph of quadratic's graph
As a general rule to solve the problem we are going to transform all values to the lower unit.
a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam
3,000 m 900 m 90 m 19 m + 7,000 m 70 m = 4,009 + 7,070 = 11,079 m
b. 5 sq.km 95 ha 8,994 sq.m + 11 sq. km. 11 ha 9,010 sq. m.
5,000,000 sq m 95,0000 sq m 8,994 sq m + 11,000,000 sq m 110,000 sq
9,010 sq m
5,103,994 sq m + 11,119,010 sq m = 16,223,004 sq m
c. 44 m – 5 dm
44 m - 0.5 m = 43.5 m
d. 73 km 47 hm 2 dam - 11 km 55 hm
73,000 m 4,700 m 20 m - 11,000 m 5,500 m
77,720 m - 16,500 m = 61,220 m
