1 revolution of a circle = circumference of that circle.
<span>1 revolution of a circle with the diameter of 28 inches = </span><span><span>πd=28π</span><span>πd=28π</span></span><span> inches. Hence, x revolutions per second = </span><span><span>28πx</span><span>28πx</span></span><span> inches per second = </span><span><span>60∗28πx</span><span>60∗28πx</span></span><span> inches </span>per minute.
<span>Given that </span><span><span>60∗28πx=35πn</span><span>60∗28πx=35πn</span></span><span> --> </span><span><span>n=<span><span>60∗28πx</span><span>35π</span></span>=48x</span><span>n=<span><span>60∗28πx</span><span>35π</span></span>=48x</span></span><span>.</span>
Answer:
<h3>B. m∠R = 110°, m∠T = 110°, m∠U = 70°</h3>
Step-by-step explanation:
The opposite angles of the parallelogram are the same.
From the diagram;
<S = <U and <R = <T
Given
<S = 70°
Since <S = <U, hence <U = 70°
Since the sum of angles in a quadrilateral is 360 degrees, hence;
<R+<S+<T+<U = 360
Since <R = <T, then;
<Y+<S+<T+<U = 360
2<T + 70+70 = 360
2<T = 360-140
2<T = 220
<T = 220/2
<T = 110°
Since <T = <R, then < R = 110°
Hence m∠R = 110°, m∠T = 110°, m∠U = 70°. Option B is correct
Answer:
10.03% probability of getting a cup weighing more than 8.64oz
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 question, we have that:

What is the probability of getting a cup weighing more than 8.64oz
This is the 1 subtracted by the pvalue of Z when X = 8.64. So

has a pvalue of 0.8997
1 - 0.8997 = 0.1003
10.03% probability of getting a cup weighing more than 8.64oz