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mojhsa [17]
2 years ago
12

Can someone please help me please

Mathematics
1 answer:
qwelly [4]2 years ago
4 0

Answer:

1) \dfrac{12+\sqrt{-16}}{4}=\dfrac{12+\sqrt{16i^{2}}}{4}\\\\\\=\dfrac{12+4i}{4}=\dfrac{12}{4}+\dfrac{4i}{4}\\\\\\= 3 + i

Real part = 3  & imaginary part = 1

2) 4 + 7i - 8 = 4 -8 + 7i

                   = -4 + 7i

Real part = -4 & imaginary part = 7

3)\dfrac{\sqrt{9}+\sqrt{-4}}{5}=\dfrac{3+\sqrt{4i^{2}}}{5}\\\\\\=\dfrac{3}{5}+\dfrac{2i}{5}\\\\\\Real \ part = \dfrac{3}{5} \ & \ imaginary \ part = \dfrac{2}{5}

4) -2-\sqrt{-25}=-2-\sqrt{25i^{2}}=-2-5i

Real part = -2  & imaginary part = -5

5)\dfrac{15-13i}{3}=\dfrac{15}{3}-\dfrac{13i}{3}\\\\\\Real \ part = \dfrac{15}{3} \ & \ imaginary \ part = \dfrac{-13}{3}

b) 1) 6 + 4i

2) -12+ 5i

3) 5-14i

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Vitek1552 [10]

Answer:

Step-by-step explanation:

Since the number of pages that this new toner can print is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

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x = the number of pages.

µ = mean

σ = standard deviation

From the information given,

µ = 2300 pages

σ = 150 pages

1)

the probability that this toner can print more than 2100 pages is expressed as

P(x > 2100) = 1 - P(x ≤ 2100)

For x = 2100,

z = (2100 - 2300)/150 = - 1.33

Looking at the normal distribution table, the probability corresponding to the z score is 0.092

P(x > 2100) = 1 - 0.092 = 0.908

2) P(x < 2200)

z = (x - µ)/σ/√n

n = 10

z = (2200 - 2300)/150/√10

z = - 100/47.43 = - 2.12

Looking at the normal distribution table, the probability corresponding to the z score is 0.017

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3) for underperforming toners, the z score corresponding to the probability value of 3%(0.03) is

- 1.88

Therefore,

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150 × - 1.88 = x - 2300

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Answer:

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Step-by-step explanation:

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Help answer this if ya know!
blagie [28]
C) The +10 is because you already have 10 coins, that will not change. The 6x is because with each passing day, the number of coins increase by 6 (x represents days). The equation then turns into y=6x+10

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Answer:

2

Step-by-step explanation:

Given the data : 29, 2, 28, 30, 26, 31

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Upper : Q3 + (1.5 * IQR)

Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range

Using calculator :

Q1 = 26

Q3 = 30

IQR = (Q3 - Q1) = 30 - 26 = 4

Lower : 26 - (1.5 * 4) = 20

Upper : 30 + (1.5 * 4) = 36

Hence, the number in the given data which falls outside the range is 2

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Answer:

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