Answer:
Sandra need to score at least <u>56%</u> in her fifth test so that her average is 80%.
Step-by-step explanation:
Given:
First 4 test scores = 87%, 92%, 76%,89%
Average targeted = 80%
We need to find the minimum score she needs to make on fifth test to achieve average of at least 80%.
Solution:
Let the minimum score she needs to make in fifth test be 'x'.
Total number of test = 5
Now we know that;
Average is equal to sum of all the scores in the test divided by number of test.
framing in equation form we get;

Multiplying both side by 5 we get;

Subtracting both side by 344 we get;

Hence Sandra need to score at least <u>56%</u> in her fifth test so that her average is 80%.
Answer:
-15
Step-by-step explanation:
Answer:
888.89 km
Step-by-step explanation:
Write the given question as an equation and solve.
80 km = 0.09 × (what distance)
(80 km)/0.09 = what distance . . . . . divide by the coefficient of the variable
888.89 km ≈ what distance
80 km is 9% of 888.89 km
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x-3)^2+(y+7)^2=64\implies [x-\stackrel{h}{3}]^2+[y-(\stackrel{k}{-7})]^2=\stackrel{r}{8^2} \\\\\\ center~(3,-7)\qquad radius=8](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%7D%7B%20r%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%28x-3%29%5E2%2B%28y%2B7%29%5E2%3D64%5Cimplies%20%5Bx-%5Cstackrel%7Bh%7D%7B3%7D%5D%5E2%2B%5By-%28%5Cstackrel%7Bk%7D%7B-7%7D%29%5D%5E2%3D%5Cstackrel%7Br%7D%7B8%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Acenter~%283%2C-7%29%5Cqquad%20radius%3D8)
so, the broadcast location and range is more or less like the picture below.
Answer:
a1=12
a6=12
a5=20
a3=20
a4=15
a2=15
as=94
Step-by-step explanation:
it is asking for each of the areas
I'm assuming As stands for areas summed