Answer:
85.9 (to the nearest hundredth)
Step-by-step explanation:
1 radian = 180°/
Therefore, 1.5 radians = 1.5 x (180°/) = 270/
= 85.9 (to the nearest hundredth)
Need help Write 6/10 as a percent, decimal
2 answers:
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The coordinate of the partition c on the line segment is (1.2, -4.7)
<h3>How to determine the coordinates of the partition?</h3>The coordinates are given as:
A = (7,-2)
B = (-8,-9)
m:n = 5:8
The coordinate of the partition is calculated using:
So, we have:
Evaluate the sum and products
Evaluate the product
(x,y) = (1.2, -4.7)
Hence, the coordinate of the partition on the line segment is (1.2, -4.7)
Read more about line segment ratios at:
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Answer:
a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The mean of a population is 74 and the standard deviation is 15.
This means that
Question a:
Sample of 36 means that
This probability is 1 subtracted by the pvalue of Z when X = 78. So
By the Central Limit Theorem
has a pvalue of 0.9452
1 - 0.9452 = 0.0548
0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
Question b:
Sample of 150 means that
This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So
X = 77
has a pvalue of 0.9929
X = 71
has a pvalue of 0.0071
0.9929 - 0.0071 = 0.9858
0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c. A random sample of size 219 yielding a sample mean of less than 74.2
Sample size of 219 means that
This probability is the pvalue of Z when X = 74.2. So
has a pvalue of 0.5793
0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2