Kate can travel 41.33 miles without exceeding her limit. This problem can be solved by using y = 2.25x + 7 linear equation with the "y" variable as the total cost that Kate must pay after she has traveled with the cab and the "x" variable as Kate's traveling distance. The equation has 7 for its constant value which is the $7 flat rate. We will find 41.33 miles as the traveling distance if we substituted the total cost with 100, which is the maximum amount that can be paid by Kate for the traveling purpose.
Answer:

Gradient = -3
• Parallel lines have the same gradient, therefore gradient, m is -3

• At point (0, -4)

y intercept is -4

Answer:
x=63.256% 63.3-rounded to the nearest tenth.
Step-by-step explanation:
Hi,
There must have a bug on the site !!!
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"Hello,Answer
=1244+8+16+32+64=124
6 minutes ago"
<span>The
first term of a geometric sequence is 4 and the multiplier, or ratio,
is 2. what is the sum of the first 5 terms of the sequence?
</span>

<span />
Answer:
--- Standard deviation
Step-by-step explanation:
Given
See attachment for graph
Solving (a): Explain how the standard deviation is calculated.
<u>Start by calculating the mean</u>
To do this, we divide the sum of the products of grade and number of students by the total number of students;
i.e.

So, we have:



Next, calculate the variance using the following formula:

i.e subtract the mean from each dataset; take the squares; add up the squares; then divide the sum by the number of dataset
So, we have:



Lastly, take the square root of the variance to get the standard deviation


--- approximated
<em>Hence, the standard deviation is approximately 11.28</em>
Considering the calculated mean (i.e. 82.76), the standard deviation (i.e. 11.28) is small and this means that the grade of the students are close to the average grade.