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Leto [7]
1 year ago
14

Brendan created a connect-the-dots puzzle in the shape of a square for his little brother. He plotted dotson a coordinate plane

at (2, 3), (7, 3), (7, -2), and (2, -2). How many units long is one side of Brendan'sconnect-the-dots puzzle? If needed, create a graph to help answer the question.
Mathematics
1 answer:
WITCHER [35]1 year ago
6 0

Solution

From the graph

Alternatively

We can use distance between two point

\begin{gathered} D=\sqrt[]{(x_2-x_1)+(y_2-y_1)^2} \\ we\text{ can use any of the two coordinates} \\ (2,3)\text{ ,(7,3)} \\ x_1=2 \\ x_2=7 \\ y_1=3 \\ y_2=3 \\ D=\sqrt[]{(7-2)^2+(3-3)^2} \\  \\ D=\sqrt[]{(7-2)^2+0} \\ D=\sqrt[]{(5^2+0} \\ D=\sqrt[]{(5^2} \\ D=5 \end{gathered}

The final answer

5 units units long is one side of Brendan's connect-the-dots puzzle

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Last year, Shantell bought a car for $24,000. The current value of the car is $21,000. Find the percent decrease in the value of
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<span>Last year, Shantell bought a car for 24 000 dollars. It decreases to 21 000 dollars this current year.
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3 years ago
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}

So, the z-score of 720 seconds is given by:

\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. 
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3 years ago
What else can the radio 4/5 be written
UNO [17]
4/5 4:5 4 to5  8 to 10 8/10 8:10
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A building company bids on two large projects. The CEO believes the chance of winning the 1st is 0.6, the chance of winning the
Ksju [112]

Answer:

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A: P(1) = .6, P(2) = .5, P(1 and 2) = .3

P(1 or 2) = P(1) + P(2) - P(1 and 2) .6 + .5 - .3 = 0.8

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