Answer:
a,e
Step-by-step explanation:
let length=l
width=w
height=h
surface area=2(l+w)×h+2×l×w
=2[lh+wh+lw]
2(8×4+4×3+8×3)=2(32+12+24)=2×68=136<160
2(7×6+6×4+7×4)=2(42+24+28)=2×94=188>160
2(3×7+7×8+3×8)=2(21+56+24)=2(101)=202>160
2(3×6+6×7+3×7)=2(18+42+21)=2(81)=162>160
2(3×5+5×7+3×7)=2(15+35+21)=2(71)=154<160
1.C) -4a-8-a = -5a-8
2. C)-2 - x - 4.3= -6.3 -x
3.A) -2x+6-x-4= -3x+2
hope it helps
Complete question :
According to the National Association of Realtors, it took an average of three weeks to sell a home in 2017. Suppose data for the sale of 39 randomly selected homes sold in Greene County, Ohio, in 2017 showed a sample mean of 3.6 weeks with a sample standard deviation of 2 weeks. Conduct a hypothesis test to determine whether the number of weeks until a house sold in Greene County differed from the national average in 2017. Useα = 0.05for the level of significance, and state your conclusion
Answer:
H0 : μ = 3
H1 : μ ≠ 3
Test statistic = 1.897
Pvalue = 0.0653
fail to reject the Null ; Hence, we conclude that their is no significant to accept the claim that number I weeks taken to sell a house differs.
Step-by-step explanation:
Given :
Sample size, n = 40
Sample mean, x = 3.6
Population mean, μ = 3
Standard deviation, s = 2
The hypothesis :
H0 : μ = 3
H1 : μ ≠ 3
The test statistic :
(xbar - μ) ÷ (s/√n)
(3.6 - 3) / (2/√40)
0.6 / 0.3162277
Test statistic = 1.897
Using T test, we can obtain the Pvalue from the Test statistic value obtained :
df = n - 1; 40 - 1 = 39
Pvalue(1.897, 39) = 0.0653
Decison region :
If Pvalue ≤ α ; Reject the null, if otherwise fail to reject the Null.
α = 0.05
Pvalue > α ; We fail to reject the Null ; Hence, we conclude that their is no significant to accept the claim that number I weeks taken to sell a house differs.
Answer:
5.673091 x 10^6
Step-by-step explanation:
