Answer:
B. 10
Step-by-step explanation:
To find the answer we would plug each given number into x to see if it makes the equation true
<em>A. 2(5) = 20 </em>
<em> 10 </em><em> 20</em>
<em>B. 2(10) = 20</em>
<em> 20 = 20</em>
<em>C. 2(20) = 20</em>
<em> 40 </em><em>20</em>
As you can see, the only answer that makes the equation 2x=20 true, is answer B
For this case we must find the inverse of the following function:
We follow the steps below:
Replace f(x) with y:
We exchange the variables:
We solve for "y":
Multiply by -2 on both sides of the equation:
We raise both sides of the equation to the square to eliminate the radical:
We subtract 3 from both sides of the equation:
We change y by f ^ {- 1} (x):
Answer:
see attached picture for #6
10 the inequality would be :
12 inch box is x
15 inch box is y
12x+15y≤84
Answer:
Step-by-step explanation:
Hello!
There are two variables of interest:
X₁: number of college freshmen that carry a credit card balance.
n₁= 1000
p'₁= 0.37
X₂: number of college seniors that carry a credit card balance.
n₂= 1000
p'₂= 0.48
a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.
The formula for the interval is:
p'₁±
0.37±1.648*
0.37±1.648*0.015
[0.35;0.39]
With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.
b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648
The formula for this interval is
p'₂±
0.48±1.648*
0.48±1.648*0.016
[0.45;0.51]
With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.
c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.
Freshmen:
Seniors:
The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.