The test scores for a math class are shown below.

Mathematics

1 answer:

Nady [450]3 years ago

6 0

Answer:

c. 4.7

Step-by-step explanation:

first, you find the mean of data set:

83 + 85 + 82 + 93 + 83 + 84 + 95 + 87 + 86 + 94 = 872/10 = 87.2

then, you take each number and subtract the mean and square it:

(87.2 - 83)² = (4.2)² = 17.64

(87.2 - 85)² = (2.2)² = 4.84

(87.2 - 82)² = (5.2)² = 27.04

(93 - 87.2)² = (5.8)² = 33.64

(87.2 - 83)² = (4.2)² = 17.64

(87.2 - 84)² = (3.2)² = 10.24

(95 - 87.2)² = (7.8)² = 60.84

(87.2 - 87)² = (0.2)² = 0.04

(87.2 - 86)² = (1.2)² = 1.44

(94 - 87.2)² = (6.8)² = 46.24

now, find the mean of the squared numbers:

17.64 + 4.84 + 27.04 + 33.64 + 17.64 + 10.24 + 60.84 + 0.04 + 1.44 + 46.24 = 219.6/10 = 21.96

then, take the square root of the number you just got:

√21.96 = 4.68

and last but not least, round to the nearest tenth: 4.7

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Solve: a + 4 = 48<br> a = 43<br> a = 51<br> a = 44<br> a = 52

andrew11 [14]

Answer:

a = 44 is correct

Step-by-step explanation:

a + 4 = 48

plug in the #'s 1 at a time

43 + 4 = 48

47 = 48 FALSE

51 + 4 = 48

55 = 48 FALSE

44 + 4 = 48 TRUE

52 + 4 = 48 FALSE

Hope this Helps!!!

5 0

3 years ago

The area of a rectangle is 27 m^2 , and the length of the rectangle is 3 m less than twice the width. Find the dimensions of the

vagabundo [1.1K]

Let $L$ be the length of the rectangle and $W$ be the width. In the problem it is given that $L=2W-3$ . It is also given that the area $LW=27$ . Substituting in the length in terms of width, we have $W(2W-3)=27 \\ 2W^2-3W-27=0 \\ (2W-9)(W+3)=0$ . Using the zero product property, $2W-9=0 \text{ or } W+3=0$ . Solving these we get the width $W=4.5 \text{ or } -3$ . However, it doesn't make sense for the width to be negative, so the width must be $\boxed{4.5 \text{ m}}$ . From that we can tell the length $L=2(4.5)-3=\boxed{6 \text{ m}}$ .