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deff fn [24]
3 years ago
12

Someone please help me❤️ Urgently with at least 1

Mathematics
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

See step-by-step explanation

Step-by-step explanation:

Question 3: (4, -3)

Solving through elimination: Multiply the first equation by -3.

-3x+6y=-30

3x-5y = 27

Add. The x drops out.

y= -3

Substitute -3 in for y.

x-2y=10

x-2(-3)=10

x+6=10

x=4

Question 4: (2,5)

Solve through elimination: Multiply the second equation with 3.

\frac{3}{2} x+y=8

\frac{9}{4} x-y=-\frac{1}{2}

3/2= 6/4

15/4x=7.5

x=2

3/2 (2) + y = 8

3+ y =8

y=5

Question 5: (6,4)

Use your graph to graph.

y=-\frac{1}{3} x+6

Slope of -1/3

Y-intercept of 6 (0,6)

X-intercept of 18 (18,0)

y=\frac{3}{2} x-5

Slope of 3/2

Y-intercept of -5 (0,-5)

X-intercept of \frac{10}{3} (

Question 6: (5, -7)

Ax+2By=-41

4Ax-By=88

3A+2B(4)=-41

4A(3)-B(4)=88

Solve by substitution:

Words of encouragement:

Good luck!

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The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds
Bumek [7]
Answer: 11.5% 

Explanation:


Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds. 

Now, we get the z-score for 720 seconds by the following formula:

\text{z-score} =  \frac{x - \mu}{\sigma}

where 

t = \text{time for the auditor to finish his work } = 720 \text{ seconds}&#10;\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}&#10;\\ \sigma = \text{standard deviation } = 100 \text{ seconds}

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\text{z-score} = \frac{x - \mu}{\sigma}&#10;\\&#10;\\ \text{z-score} = \frac{720 - 600}{100}&#10;\\&#10;\\ \boxed{\text{z-score} = 1.2}

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

P(t \ \textgreater \  720) &#10;\\ = P(z \ \textgreater \  1.2)&#10;\\ = 1 - P(z \leq 1.2)&#10;\\ = 1 - 0.885&#10;\\  \boxed{P(t \ \textgreater \  720)  = 0.115}

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice. 
8 0
2 years ago
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