Solution:
we are given that
John is 3 years older than Jim. Jim is 4 years less than half of Dana’s age.
Let the age of Jim be x years then we can say
Age of John
If diana Age is D then

It can be re-wriiten as
Diana age 
Given that their ages add up to 167

Hence jim is 39 nyears old, John is 39+3=42 years old. and Diana is 2*39+8=78+8=86 years old.
Answer:
x=17
Step-by-step explanation:
3x-24=x+10
Carry over the like terms and put them together. Make the unknowns which are x the subject.
3x-x= 24+10
2x= 34
2x= 34/2
x= 17
Answer:
H0:p1=p2; Ha:p1≠p2, which is a two-tailed test.
Step-by-step explanation:
We formulate our hypotheses as
H0:p1=p2; Ha:p1≠p2, which is a two-tailed test.
Supposing the probability or proportion of the first survey is equal to the probability or proportion of the second survey. This will be the null hypothesis and the alternative hypotheses would be that these two proportions or probabilities are unequal.
This is a two tailed test.
Answer: 
<u>Step-by-step explanation:</u>

Answer:
3
Step-by-step explanation:
(6 + 2 x 6)/(15 - 11 + 2)
(6 + 12)/(4 + 2)
18/6
3