Find if the triangles are similar. Please show your work, I need it done by tomorrow morning
2 answers:
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Using the <u>normal distribution and the central limit theorem</u>, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem:
- The mean is of
.
- The standard deviation is of
.
- Sample of 100, hence
The interval that contains 95.44% of the sample means for male students is <u>between Z = -2 and Z = 2</u>, as the subtraction of their p-values is 0.9544, hence:
Z = -2:
By the Central Limit Theorem
Z = 2:
The interval that contains 99.44% of the sample means for male students is (3.4, 3.6).
You can learn more about the <u>normal distribution and the central limit theorem</u> at brainly.com/question/24663213
To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.