0.7=7/10
0.8=8/10
0.75=3/4 this is between 0.7 and 0.8
Answer:
Step-by-step explanation:
Hello!
To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:
Sample 1: Compressed soil
X₁: penetrability of a plot with compressed soil.
n₁= 20 plots
X[bar]₁= 2.90
S₁= 0.14
Sample 2: Intermediate soil
X₂: penetrability of a plot with intermediate soil.
n₂= 20 (with outlier) n₂= 19 plots (without outlier)
X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)
S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)
Outlier: 4.26
Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:
Parámeter of interest: μ₁-μ₂
Interval:
[(X[bar]₁-X[bar]₂)±
*Sa
]


[(2.90-2.29)±2.715*0.20
]
[0.436; 0.784]
I hope this helps!
Hi there ! \(๑╹◡╹๑)ノ♬
Answer: 
Step-by-step explanation:





Answer:
An equation
Step-by-step explanation:
this can be solved too
4g + 7e + 4 + 9 + (-4e) + 11 = 5g + 3e + 15
combine like terms
4g + 3e + 24 = 5g + 3e + 15
3e + 24 = g + 3e + 15
24 = g + 15
9 = g
and
0 = e or *no solution* since e cancels itself out.
Answer:
Size, angle measures, position of center of rotation, shape.