Answer: independent variable: time (t). Dependent variable: distance (d).
Step-by-step explanation: an independent variable is a variable whose variation does not depend on that of another. In the given problem, the independent variable is time, because time will pass by no matter what (without depending in any other variable). The dependent variable in this case is the distance, because how far Jillian goes, depends on how much time she expends walking and jogging.
58% is about equal to 60%, 10% of 121 is 12.1.
12.1 * 6 = 72.6
To take this further, 58% is 2% less than 60%, or 2 times 1%. 1% of 121 is 1.21.
72.6 - (1.21 * 2) = 72.6 - 2.42 = 70.18
Answer:
The image of
through T is ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We know that
→
is a linear transformation that maps
into
⇒

And also maps
into
⇒

We need to find the image of the vector ![\left[\begin{array}{c}4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D)
We know that exists a matrix A from
(because of how T was defined) such that :
for all x ∈ 
We can find the matrix A by applying T to a base of the domain (
).
Notice that we have that data :
{
}
Being
the cannonic base of 
The following step is to put the images from the vectors of the base into the columns of the new matrix A :
(Data of the problem)
(Data of the problem)
Writing the matrix A :
![A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now with the matrix A we can find the image of
such as :
⇒
![T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
We found out that the image of
through T is the vector ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
Answer:
9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated
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:

What is the probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated?
This is 1 subtracted by the pvalue of Z when X = 16. So



has a pvalue of 0.9082
1 - 0.9082 = 0.0918
9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated