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AleksAgata [21]
3 years ago
14

Complete this equation 2 2/3 +4/5= 40/? + 12/? = ?/?

Mathematics
1 answer:
OlgaM077 [116]3 years ago
6 0

Answer:(7/8 - 4/5)^2 = 9

1600

= 0.005625

Step-by-step explanation:

Subtract: 7

8

- 4

5

= 7 · 5

8 · 5

- 4 · 8

5 · 8

= 35

40

- 32

40

= 35 - 32

40

= 3

40

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(8, 5) = 40. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 8 × 5 = 40. In the next intermediate step the fraction result cannot be further simplified by cancelling.

In words - seven eighths minus four fifths = three fortieths.

Exponentiation: the result of step No. 1 ^ 2 = (3

40

) ^ 2 = 32

402

= 9

1600

In words - three fortieths squared = nine one-thousand six-hundredths.

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step 1

Find the measure of angle 1

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so

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90+58+<1=180

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<1=180-148

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step 2

Find the measure of angle 2

we know that

<2=<1 -----> bt vertical angles

so

<2=32 degrees

step 3

Find the measure of angle 3

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In the triangle CDE

32+108+<3=180

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Answer:

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Suppose that scores on an aptitude test are normally distributed with a mean of 100 and a standard deviation of 4.6. Scores on a
DerKrebs [107]

Answer:

<u>The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

Step-by-step explanation:

1. Let's check all the information given to us to answer the question correctly:

Mean of the scores on the aptitude test = 100

Standard deviation of the aptitude test = 4.6

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2. Which statement best describes Felix's scores on the two tests comparatively?

Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.

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Now, we calculate the z-score, using the value of the standard deviation this way:

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