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borishaifa [10]
2 years ago
9

I didnt put the answer choices because ppl be putting random stuff

Mathematics
1 answer:
lesya692 [45]2 years ago
5 0
I’m pretty sure the answer is 127.48 ,but I could be wrong.
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Add them all together and there is your sum
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3 years ago
It takes a guard 3 hours to walk 12 miles on foot. Another guard travels by horse at a rate of 14 miles per hour. How long will
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Answer:

around 50 minutes

Step-by-step explanation:

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3 years ago
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Nanu borrowed a certain sum at the rate of 10 % p.a. If she paid compound interest Rs. 1,290 at the end of two years compounded
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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
Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
tankabanditka [31]
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}&#10;b=\textit{rate of the boat}\\&#10;r=\textit{rate of the river}&#10;\end{cases}\qquad thus&#10;\\\\\\&#10;&#10;\begin{array}{lccclll}&#10;&distance&rate&time(hrs)\\&#10;&----&----&----\\&#10;upstream&48&b-r&4\\&#10;downstream&48&b+4&3&#10;\end{array}&#10;\\\\\\&#10;&#10;\begin{cases}&#10;48=(b-r)(4)\to 48=4b-4r\\\\&#10;\frac{48-4b}{-4}=r\\&#10;--------------\\&#10;48=(b+r)(3)\\&#10;-----------------------------\\\\&#10;thus\\\\&#10;48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)&#10;\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
5 0
3 years ago
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