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Simora [160]
2 years ago
9

Candace had scores of 94, 83, 92, and 86 on her first four exams of the semester. What score must she obtain on the fifth exam t

o have an average of 90 or better for the five exams?
Mathematics
2 answers:
USPshnik [31]2 years ago
3 0

Answer:

A score of 95 she must  obtain on the fifth exam to have an average of 90 or better for the five exams

Step-by-step explanation:

average = sum of all terms / number of terms

average = 94 + 83 + 92 + 86 + x / 5

90 = 355 + x / 5

multiply both sides by 5,

90 * 5 = 355 + x / 5 * 5

450 = 355 + x

subtract 355 from both sides to get x alone

450 - 355 = 355 + x - 355

95 = x

e-lub [12.9K]2 years ago
3 0

Answer:

  95

Step-by-step explanation:

The total of deviations from average must be zero.

<h3>deviations</h3>

  (94 -90) = 4

  (83 -90) = -7

  (92 -90) = 2

  (86 -90) = -4

and for the 5th exam:

  (x -90) . . . . where the exam score is x

__

<h3>total of deviations</h3>

We want the sum to be zero:

  4 -7 +2 -4 +(x -90) = 0

  x -95 = 0 . . . . . . . simplify

  x = 95 . . . . . . . . . add 95

Candace must get a 95 or better to have an average of 90 for the 5 exams.

_____

<em>Additional comment</em>

When we're figuring this out mentally, we observe that 4 and -4 cancel, so Candace needs 5 more points than average to balance the net -5 she has from (-7+2). That is, she needs 90+5 = 95.

For many problems involving averages or sequences of numbers, it is often convenient to look at the deviations from average.

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3 years ago
The measures of two supplementary angles are (2x – 8)° and (3x - 2)°. What is the measure of the
saw5 [17]

Answer:

68°

Step-by-step explanation:

Here,

  • Two supplementary angles are (2x – 8)° and (3x - 2)°.

As we know that the sum of two supplementary angles are 180°. So,

→ (2x – 8)° + (3x – 2)° = 180°

→ 2x° – 8° + 3x° – 2° = 180°

→ 5x° – 10° = 180°

→ 5x° = 180° + 10°

→ 5x° = 190°

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→ <u>x</u><u> </u><u>=</u><u> </u><u>3</u><u>8</u><u>°</u>

Supplementary angles are,

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4 0
3 years ago
An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t)
beks73 [17]
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