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Oksana_A [137]
2 years ago
11

During his 360-min. shift at the local grocery store, Jared spent 122 min. stocking shelves, 74 min. bagging groceries, and 62 m

in. sweeping the floors. He also took a 35-min. lunch break. Use front-end estimation to find about how many minutes Jared had left for miscellaneous tasks.
A. 70
B. 80
C. 90
D. 100
Mathematics
1 answer:
sergeinik [125]2 years ago
3 0

Answer:

d.100

Step-by-step explanation:

i just took the test

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To better give a visual, I drew it out instead.

The attachment is the solution.

8 0
1 year ago
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The distance between city E and F is 678.35 miles and the distance between city F and G is 156.8 miles. What is the distance bet
Free_Kalibri [48]

E to F = 678.35 miles
F to G = 156.8 miles
G to H = x
Total distance = 2,457 miles
E to G = 678.35 + 156.8 miles
E to G = 835.15
G to H = 2,457 - 835.15
G to H = 1621.85 miles

8 0
3 years ago
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I’m not sure how to do this please help
Nikolay [14]

Answer:

B :step 2   she didnt collect all the like terms and calculate

Step-by-step explanation:

first rewrite  remove all (  )Parentheses

2nd   collect all like terms and calculate

a^4 + 7a -16 -12^a^3 + 5a -3

a^4 + 12a - 19 - 12a^3     ( Like terms are 7a +5a  and -16 +-3)

so she skipped the second step

3 0
2 years ago
Name the identical triangles and measurement of side BA
aalyn [17]

Step-by-step explanation:

BE/BC=BD/BA

tHE TWO SIMILAR TRIANGLES ARE BED AND BC. tHE PROPORTION ABOVE REPRESENTS USE OF THE TWO SIMILAR TRIANGLES.

4/(4 + 10) = 5/(5+X) Cross Multiply

4(5+x) = 5(4+10)m

20 +4x=20 +50 subtract 20 from both sides

4x = 50 divide by 4

x =12.5

4 0
2 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:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

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

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
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