If a rectangle has a width of (2 x 10^4) cm and a length of (8 x 10^7)cm, what is the area of the rectangle ,expressed in scient
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The ratio of the side measures are 6:7:9.
Let's assume length of three sides of the triangle is 6x, 7x and 9x.
Given,perimeter of a triangle is 110 inches.. Perimeter is the sum of all three sides of the triangle. So, we can set up an equation as follows:
6x+7x+9x= 110
22x =110
Dividing each sides by 22.
x=5.
Now we can plug in x=5 in all the expressions of sides.
So, 6x=6(5)=30,
7x=7(5)=35
9x=9(5)=45.
So, the sides of the triangle are 30 inches, 35 inches and 45 inches.
Answer:
P(111.2-cm < ¯ x < 111.4-cm) = 0.4726
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 normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, 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.
In this problem, we have that:
Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm.
This is the pvalue of Z when X = 111.4 subtracted by the pvalue of Z when X = 111.2. So
X = 111.4
By the Central Limit Theorem
has a pvalue of 0.5.
X = 111.2
has a pvalue of 0.0274.
0.5 - 0.0274 = 0.4726
P(111.2-cm < ¯ x < 111.4-cm) = 0.4726