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aniked [119]
3 years ago
13

How tall is the flagpole??

Mathematics
2 answers:
allochka39001 [22]3 years ago
6 0

Answe25 feet tall

Step-by-step explanation:

ser-zykov [4K]3 years ago
6 0

Answer:

I would assume 186

Step-by-step explanation:

Since 48x3=144

62x3=186

Or x

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What fraction of 130 is 80
azamat
80/130 simplified would just be 8/13, and that is the simplest form.
Final answer: 8/13
Hope this helped! :)
5 0
3 years ago
A paper has a diameter of 9 inches what is the circumference of the plate
wlad13 [49]

The answer will be

c = \pi \: d \\ c = 9\pi \\  \\ c = 28.26in

5 0
4 years ago
How to solve 4x+6y less than or equal to 120
yaroslaw [1]
120 - 4 - 6 = ( that is the process, you ca find the answer)
6 0
2 years ago
Mandy and Jeff start running around the track in the same direction at the same time. Mandy runs one lap in 2 minutes and 30 sec
Savatey [412]

Answer:

At 8:52:30 am they will be side-by-side again

Step-by-step explanation:

Let s be the distance ran in one 1 lap.

Here Jeff runs one extra lap compared to Mandy for coming side by side again.

Mandy runs one lap in 2 minutes and 30 seconds = 150 s

Speed of Mandy, =\frac{s}{150}

Jeff runs one lap in 2 minutes and 15 seconds = 135 s

Speed of Jeff, =\frac{s}{135}

Let t be the time when Jeff run one extra lap,

That is

              t\times \frac{s}{135}-t\times \frac{s}{150}=s\\\\t\left ( \frac{1}{135}-\frac{1}{150}\right )=1\\\\t\times \frac{15}{135\times 150}=1\\\\t=1350s

So after 1350 seconds Jeff comes side by side to Mandy,

They both start at 8:30 am

             1350 s = 22.5 minutes = 22 minutes 30 seconds

          Time = 8:52:30 am

3 0
3 years ago
A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 20%
zhuklara [117]

Answer:

a. 5% of the employees will experience lost-time accidents in both years

b. 24% of the employees will suffer at least one lost-time accident over the two-year period

Step-by-step explanation:

a. What percentage of the employees will experience lost-time accidents in both years?

20% last year, of those who suffered last year, 25% during this year. So

p = 0.2*0.25 = 0.05

5% of the employees will experience lost-time accidents in both years.

b. What percentage of the employees will suffer at least one lost-time accident over the two-year period?

5% during the two years.

10% during the current year. 25% of employees who had lost-time accidents last year will experience a lost-time accident during the current year.

So the 10% is composed of 5% during both years(25% of 20%) and 5% of the 80% who did not suffer during the first year.

First year yes, not on the second.

75% of 20%. So, total:

0.05 + 0.05*0.8 + 0.75*0.2 = 0.24

24% of the employees will suffer at least one lost-time accident over the two-year period

3 0
3 years ago
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