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allochka39001 [22]
3 years ago
15

Jacki evaluated the expression below. 2 cubed (3 minus 1) + 4 (8 minus 12) = 2 cubed (2) + 4 (4) = 8 (2) + 16 = 16 + 16 = 32. Wh

at was Jacki’s error? Jacki should not have multiplied 8 and 2. Jacki should have multiplied 4 and 8 first. Jacki should have simplified the exponent first. Jacki did not subtract 12 from 8 correctly.
Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
5 0

Answer:

Your anwser would be D

Jacki did not subtract 12 from 8 correctly.

Step-by-step explanation:

It should be a negative instead of a positive

ziro4ka [17]3 years ago
4 0

Answer:

its c

Step-by-step explanation:

did the test

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Simplify the following expression below:
IrinaK [193]
So, 4/3 - 2i

4/3 - 2i = 12/13 + i8/13

multiply by the conjugate: 
3 + 2i/3 + 2i
= 4(3 + 2i)/(3 - 2i) (3 + 2i)
(3 - 2i) (3 + 2i) = 13

(3 - 2i) (3 + 2i)

apply complex arithmetic rule: (a + bi) (a - bi) = a^2 + b^2
a = 3, b =  - 2
= 3^2 + (- 2)^2

refine: = 13
= 4(3 + 2i)/13

distribute parentheses:
a(b + c) = ab + ac
a = 4, b = 3, c = 2i
= 4(3) + 4(2i)

Simplify:
4(3) + 4(2i)
12    +    8i


4(3) + 4(2i)
Multiply the numbers: 4(3) = 12
= 12 + 2(4i)

Multiply the numbers: 4(2) = 8
= 12 + 8i


   12 + 8i
= 12 + 8i/13

Group the real par, and the imaginary part of the complex numbers:


Your answer is: 12/13 + 8i/13





Hope that helps!!!
7 0
3 years ago
Will give 25 points pleaseee help
Sonja [21]

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Step-by-step explanation:

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Step-by-step explanation:

7 0
3 years ago
he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s
sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

P(X

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

NORMSDIST(6.366)-NORMSDIST(-1.47)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
3 years ago
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