In a class in which the final course grade depends entirely on the average of four equally weighted 100-point tests, Erica has scored 86, 79, and 85 on the first three. What range of scores on the fourth test will give Erica a C for the semester (an average between 70 and 79, inclusive)? Assume that all test scores have a non-negative value.

Erica must score between 30 and 66 to obtain a grade average of C.  

Step-by-step explanation:

We are given the following ion the question:

Erica's score:

86, 79, and 85

To get a grade C, the average score of Erica must lie between 70 and 79.

Formula:

Let x be the score on fourth test.

Thus, we can write:

Thus, Erica must score between 30 and 66 to obtain a grade average of C.

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier