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Marizza181 [45]
4 years ago
8

Estimate the product then complete the multiplication

Mathematics
1 answer:
kvv77 [185]4 years ago
7 0
Since 11/4 is close to 12, the estimate would be 12 x 8.
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Edger needs 6 cookies and 2 brownies for every 4 plates how many cookies and brownies does he need for 10 plates
evablogger [386]

Answer:

Edger needs 15 cookies and 5 brownies.

Step-By-Step:

6 + 6 + 3 = 15

2 + 2 + 1 = 5

4 can go into 10, 2.5 times

6 0
3 years ago
Use substitution. <br> 2x +3y = -5<br> 4x+3y= 17
Goshia [24]

Answer:

Step-by-step explanation:

-2x -3y = 5

4x + 3y = 17

2x = 22

x = 11

4(2) + 3y = 17

8 + 3y = 17

3y = 9

y = 3

(11,3)

3 0
3 years ago
Read 2 more answers
Alana took a math quiz last week. She got 12 out of 20 problems correct. What percentage did Alana get correct?
Deffense [45]

Answer:

60%

Step-by-step explanation:

The fraction 12/20 in simplest form would be 6/10 which come be converted to a percentage. (6x10)/(10x10)=60/100 or 60%.

4 0
3 years ago
How do I make a formula for a line if they only give me the intercept and the slope
Eddi Din [679]
Well what you want is y = mx + b
m = slope
b = intercept for y
with that just substitute in the slope and the intercept, but leave the x and y as it is. So later on you can put in a x coordinate and get the y coordinate through that, hope this helps!
8 0
3 years ago
Read 2 more answers
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. 
8 0
3 years ago
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