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sergey [27]
2 years ago
6

Sharon’s turtle escaped from her backyard sometime in the last few hours. According to her calculations, the farthest the turtle

could have gone is 4 blocks down the road in either direction. If Sharon lives on the 112th block of town, which equation can be used to find the block numbers that represent the farthest distance that the turtle may be?]
Mathematics
1 answer:
laila [671]2 years ago
6 0

Turtle can go in any direction farthest  to 4 block

hence Maximum Distance can be + 4 or - 4

Block representing farthest distance is x

Distance of block  x from 112  would be

x  - 112  

x  - 112    = ± 4

Taking mod both sides

=> | x - 112 | = |  ± 4 |

=> | x - 112 | = 4

| x - 112 | = 4  is the equation which can be used to find the block numbers that represent the farthest distance that the turtle may be

Brainliest?

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Multiply the amount she puts in the account every week by the number of weeks (w) and add that to what she already has saved:

20w + 600 >= 2000

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3 years ago
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A dog sled race is 25 miles long.The equation 5/8 k=25 can be used to estimate the race’s length k in kilometers. Approximately
EastWind [94]
The distance to be covered in the race is 25 miles
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1 mile=0.621371 km
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Len [333]

Answer:

96

Step-by-step explanation:

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8 0
3 years ago
1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

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It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

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P( 70

F) More than 130

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8 0
3 years ago
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