Answer:
Mr.Lin's present age is 30 years now. His son is 8 years old now.
Step-by-step explanation:
Let L = Mr.Lin's present age,
S = his son's present age.
First equation is
L + S = 38. (1)
The second equation is
L + 3 = 3*(S+3) (2) ("In three years' time, Mr.Lin will be three times as old as his son.")
From (1), express S = 38-L and substitute it into (2). You will get
L + 3 = 3*((38-L)+3), or
L + 3 = 114 - 3L + 9,
4L = 114 + 9 - 3,
4L = 120,
L = 30.
The answer is Johnny. You can conclude from the first sentence of your question. It is stating that Johnny's mother has three children. Johnny is included in the three children. So, the third child's name was Johnny!
Thank you! :D
You can use the distance formula
Sqroot((x2-x1)^2 + (y2-y1)^2)
(2,-1) and (5,3), use given points
Sqroot((5-2)^2 + (3-(-1))^2)
Sqroot((3)^2 + (4^2)
Sqroot(9+16) = squareroot of 25
Squareroot of 25 = 5
Solution: D. 5 units
Answer:
And on this case if we see the significance level given 
we see that 
so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.
Step-by-step explanation:
A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.
represent the p value for the test
O= obserbed values
E= expected values
The system of hypothesis for this case are:
Null hypothesis: ![O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i](https://tex.z-dn.net/?f=O_i%20%3D%20E_i%5B%2Ftex%5B%3C%2Fp%3E%3Cp%3EAlternative%20hypothesis%3A%20%5Btex%5DO_i%20%5Cneq%20E_i%20)
The statistic to check the hypothesis is given by:
On this case after calculate the statistic they got: 
And in order to calculate the p value we need to find first the degrees of freedom given by:
, where k represent the number of levels (on this cas we have 10 categories)
And in order to calculate the p value we need to calculate the following probability:
And on this case if we see the significance level given we see that so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.
