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Shkiper50 [21]
3 years ago
11

Simplify the expression to a + bi form: (2 – i) – (-4+8i)

Mathematics
2 answers:
Ne4ueva [31]3 years ago
8 0

Answer:

6-9i

Step-by-step explanation:

(2-i)-(-4+8i)

2-i+4-8i

2+4-i-8i

6-i-8i

6-9i

If you're satisfied with the answer, then please mark me as Brainliest.

ser-zykov [4K]3 years ago
5 0

Answer:

→(2 - i) - ( - 4 + 8i) \\  = 2 - i + 4 - 8i \\  = 2 + 4 - i - 8i \\  =  \boxed{6 - 9i}✓

  • <u>6 - 9i</u> is the right answer.
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8x^3y^4(3 sqrt xy) LETTER A :)

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Helppppppppppppppppp asappppppp
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Answer:

140

Step-by-step explanation:

<u>As per picture:</u>

∠PTQ  and ∠RTS are vertical angles and they are equal

  • ∠PTQ = (x + 28)°
  • ∠RTS = (2x + 16)°
  • ∠PTQ  = ∠RTS
  • x + 28 = 2x + 16
  • 2x - x = 28 - 16
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∠PTR is supplementary with ∠PTQ  and their sum is 180°

  • ∠PTQ = 12 + 28 = 40°
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3 years ago
two-eighths of the students are boys. If there are 13 boys in the class, how many students are there in all?​
EleoNora [17]

Answer:

52 students

Step-by-step explanation:

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6.5 times 8 is 52

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3 years ago
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2/5 y + 1/5 x - 0.26- 6+ (-2)
Lina20 [59]

Answer:

Simplified

2/5y + 1/5x - 413/50

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Step-by-step explanation:

8 0
3 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev
Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that \mu = 201.9, \sigma = 2.1.

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7

This probability is 1 subtracted by the pvalue of Z when X = 204.1.

Z = \frac{X - \mu}{\sigma}

Z = \frac{204.1 - 201.9}{0.7}

Z = 3.14

Z = 3.14 has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0
3 years ago
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