Answer:
27/462
OR
.058
Step-by-step explanation:
the probability that the first student, selected at random, is a nursing student is 3/22.
the probability that the second student, selected at random from the remaining students, is an accounting major is 9/21.
the probability the first student chosen is a nursing student and the second student chosen is an accounting major is therefore 3/22 * 9/21 = 27/462.
Answer:
i just need points
Step-by-step explanation:
<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>
<h2>Please Mark me as brainlist. </h2>
Easy, so absolute value makes anything inside into positive
|x|=2
therefor x=-2 or x=2 since it would be made posotive
answer is -2 and 2
Answer:
50%
Step-by-step explanation:
68-95-99.7 rule
68% of all values lie within the 1 standard deviation from mean 
95% of all values lie within the 1 standard deviation from mean 
99.7% of all values lie within the 1 standard deviation from mean 
The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4.
68% of all values lie within the 1 standard deviation from mean = 
= 
95% of all values lie within the 2 standard deviation from mean = 
= 
99.7% of all values lie within the 3 standard deviation from mean = 
= 
Refer the attached figure
P(43<x<55)=2.5%+13.5%+34%=50%
Hence The approximate percentage of light bulb replacement requests numbering between 43 and 55 is 50%
