Answer:A quadrilateral always has a sum of 360o in total.
Answer:
The mean birth weight for the sampling distribution is
3,500 grams.
Step-by-step explanation:
The sample mean is the average of the sample values collected divided by the number of the samples, while the population mean is the average or mean of all the values in the population. If the sample is random and the sample size is large enough, then the sample mean would be a good estimator of the population mean. This implies that with a randomly distributed and unbiased sample size, the sample mean and population mean will be equal, according to the central limit theorem. Therefore, the mean of the sample means will always approximate the population mean.
Answer:
Step-by-step explanation:
given ,
equation y=ax²+bx+c
passing through points (1,6), (3, 20), and (−2,15).
then these points will satisfy the equation
at (1,6)
y = a x²+b x+c
6 = a(1)² + b (1) + c
a + b + c = 6------(1)
at (3 , 20)
y = a x²+b x+c
20 = a(3)² + b (3) + c
9 a + 3 b + c = 20------(2)
at (−2,15)
y = a x²+b x+c
15 = a(-2)² + b (-2) + c
4 a -2 b + c = 15------(3)
solving equation (1),(2) and (3)
a = 6 - b - c
9 (6 - b - c)+ 3 b + c = 20
6 b + 7 c = 34-------(4)
4 (6 - b - c) -2 b + c = 15
2 b + c = 3----------(5)
on solving equation (4) and (5)
Answer:
the prices were $0.05 and $1.05
Step-by-step explanation:
Let 'a' and 'b' represent the costs of the two sodas. The given relations are ...
a + b = 1.10 . . . . the total cost of the sodas was $1.10
a - b = 1.00 . . . . one soda costs $1.00 more than the other one
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Adding these two equations, we get ...
2a = 2.10
a = 1.05 . . . . . divide by 2
1.05 -b = 1.00 . . . . . substitute for a in the second equation
1.05 -1.00 = b = 0.05 . . . add b-1 to both sides
The prices of the two sodas were $0.05 and $1.05.
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<em>Additional comment</em>
This is a "sum and difference" problem, in which you are given the sum and the difference of two values. As we have seen here, <em>the larger value is half the sum of the sum and difference</em>: a = (1+1.10)/2 = 1.05. If we were to subtract one equation from the other, we would find <em>the smaller value is half the difference of the sum and difference</em>: b = (1.05 -1.00)/2 = 0.05.
This result is the general solution to sum and difference problems.
Answer:
Let the G represent the Girl child
and, B represents the Boy child.
Then Sample Space = {BGG, GBG, GGB, BBG, BGB, GBB, BBB, GGG}
Thus, There are 8 sample space.
Then, Probability that couple have 1 boy = {BGG, GBG, GGB}
= 3 ÷ 8 = 0.375
Probability that couple have 2 boys = { BBG, BGB, GBB}
= 3 ÷ 8 = 0.375
Probability that couple have 3 boys = {GGG}
= 1 ÷ 8 = 0.125
