Express (x-6)^2 as a trinomial in standard form
2 answers:
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The answer is 96%.
Explanation:
It is generally presumed that the scores are normally distributed.
1) You are given how many standard deviations from the mean Jeremy's score is. This is exactly the definition of the z-score. Therefore z = 1.75
2) Look at a left-tail z-table in order to find the area of the normal curve on the left of your z-score (see picture attached). A = 0.9599
3) Multiply the area by 100 in order to find the percentile:
<span>0.9599 </span>× 100 = 95.99
Therefore, 95.99% of the students scored less than Jeremy.
Hence, the answer is 96%.
Explanation:
It is generally presumed that the scores are normally distributed.
1) You are given how many standard deviations from the mean Jeremy's score is. This is exactly the definition of the z-score. Therefore z = 1.75
2) Look at a left-tail z-table in order to find the area of the normal curve on the left of your z-score (see picture attached). A = 0.9599
3) Multiply the area by 100 in order to find the percentile:
<span>0.9599 </span>× 100 = 95.99
Therefore, 95.99% of the students scored less than Jeremy.
Hence, the answer is 96%.
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!