Answer:
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.7291
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have 
. So
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
The first one means what is the smallest number that goes into 12 which would be 4. The second one mean what does not go into 4 and 6 so that would be 8 because 6 is not a multiple of 8.The third one is either 1 or 2 not sure. And the 4th one is little complicated you might want to look up video on prime factorization that can really help you .
The ball will hit the ground at 0.125 seconds
<h3>
Quadratic function</h3>
The equation of the function is a quadratic function, and the function is given as:
When the ball hits the ground, the value of h is 0.
So, the equation of the function becomes
Subtract 2t from both sides of the equation
Divide both sides by -2t
Divide both sides by 8
Hence, the ball will hit the ground at 0.125 seconds
Read more about quadratic functions at:
brainly.com/question/11631534
Start by putting two coins on each side of the scale. If one side is higher, then the light coin must be in that pair (use a second weighing to determine which of that pair is the lightest.)
If the initial weighing shows the two pairs to be even, discard those four and go to the three others. Put two of the coins on the scale - one on each side. If one side stays higher, that coin is your light one. If however they are equal, the final unweighed coin must be the lighter coin.
