<span>Part A
</span>f(x)=30∗0.2xand<span>f(x)=45+3x</span><span>
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Part B</span>
f(x)=30+0.2(5)f(x)=45+3(5)<span>
Neighborhood A and neighborhood B both have 60 houses after 5 years
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Part C
1 year
A=36
B=48
2 years
A=42
B=51
3 years
A=48
B=54
4 years
A = 54
B = 57
5 years
A = 60
B= 60
Answer:
224
Step-by-step explanation:
Multiply all the numbers
Answer:
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Step-by-step explanation:
The sample proportion is p2= 7/27= 0.259
and q2= 0.74
The sample size = n= 27
The population proportion = p1= 0.4
q1= 0.6
We formulate the null and alternate hypotheses that the new program is effective
H0: p2> p1 vs Ha: p2 ≤ p1
The test statistic is
z= p2- p1/√ p1q1/n
z= 0.259-0.4/ √0.4*0.6/27
z= -0.141/0.09428
z= -1.496
The significance level ∝ is 0.05
The critical region for one tailed test is z ≤ ± 1.645
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Answer:
AAS
Step-by-step explanation:
Answer:
subtract 27 from 85
Step-by-step explanation:
85
-27
———
58
7 cannot be subtracted from 5 so borrow a 10 to make it 7 subtracted from 15 and then the 8 tens becomes 7 tens so 7-2=5
58+27=85 to check your work
