There were 340,000 cattle placed on feed
How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?
Given the fraction of total cattle for 700 - 799 pounds is 2/5
Let x be the number of cattle between 700 - 799 pounds
We make a proportion using the fraction
Cross multiply it and solve for x
340000* 2 = 5x
680000 = 5x
Divide by 5 on both sides
So x= 136,000
There were 136,000 cattle between 700 and 799 pounds
Answer:
As per the given statement:
The region bounded by the given curves about the y-axis, , y=0, x = 0 and x = 1
Using cylindrical shell method:
The volume of solid(V) is obtained by rotating about y-axis and the region under the curve y = f(x) from a to b is;
where
where x is the radius of the cylinder
f(x) is the height of the cylinder.
From the given figure:
radius = x
height(h) =f(x) =y=
a = 0 and b = 1
So, the volume V generated by rotating the given region:
therefore, the volume of V generated by rotating the given region is
Answer:
26.11% of women in the United States will wear a size 6 or smaller
Step-by-step explanation:
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.
In this problem, we have that:
In the United States, a woman's shoe size of 6 fits feet that are 22.4 centimeters long. What percentage of women in the United States will wear a size 6 or smaller?
This is the pvalue of Z when X = 22.4. So
has a pvalue of 0.2611
26.11% of women in the United States will wear a size 6 or smaller