Given that median area is 15 square units.
Hence rectangle C in the middle has 15 square units.
Its dimensions can be width= 5 and length = 3
SInce B is smaller than C and has the same length, B has lengh of 3 with area = 9 sq units.
D has the same perimeter = 16 units. Since D is a square, side of D = 4 units.
Now D and E have the same length. Hence length of E = 4 units.
Width of E = width of C = 5 units.Thus makes the area of E as 20 sq units.
Rectangle A has length =3 and width can be less than 3 since area is smaller than B.
so A has length= 3 width = 2 with area = 6 sq units.
72 is the minimum grade he must get on the last test in order to have an average of 77.
<u>Step-by-step explanation:</u>
The grades of a student are given 72,91,78,72 and the grade of his last test is not given.
- You have to find the minimum grade the student shall get, so that the student average must be 77.
- The four grades are already given. Therefore, we need to find only the fifth grade.
The term average is defined as the sum of all the data in a set divided by the number of data in a set.
Here, the number of data is 5. (Because the students has 4 grades plus one grade for his last test).
The average he should get is 77.
Average = Sum of all grades / number of grades
Let, 'x' be the grade of the last test.
⇒ 77 = (72+91+78+72+x) / 5
⇒ 77 = (313+x) / 5
⇒ 385 = 313 + x
⇒ x = 385 - 313
⇒ x = 72
The minimum grade he must get on the last test is 72.
Answer:
#3
Step-by-step explanation:
-47 = x + 15
Rearrange it to isolate x:
-47 - 15 = x
-62 = x
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