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Alex_Xolod [135]
3 years ago
6

Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.a.

) assuming that they start at the same point ant the same time, how much sooner does the faster arrive at a destination 780 m away?b.) how far would the students have to walk so that the faster student arrives 5.50 min before the slower student?
Mathematics
1 answer:
anyanavicka [17]3 years ago
5 0

Answer:

a)The second student arrive 7 minutes and 36 seconds sooner

b)They have to walk 598,51 m

Step-by-step explanation:

Speed= Distance/Time

We are looking for time now

Time = Distance/Speed

First student

Time = 780 m/0.90 m/s=866,66 seconds

Second student

Time = 780 m/1.90 m/s=410,52 seconds

Difference=866,66-410,52= 456,14 seconds

In minutes

Difference= 456,14 seconds /60=7 minutes 36 seconds

Distance=Speed x Time

Speed¹ x Time¹ =Speed² x Time²

Time¹= Time² + 5.50 min

5.50 min =350 seconds

Time²  =Speed¹ x Time¹ /Speed²

Time²  =Speed¹ x (Time² + 350 seconds  ) /Speed²

Time²  =0.9 x (Time² + 350 seconds  ) /1.9

Time²  =0,4737 x (Time² + 350 seconds  )

Time²  =0,4737 Time² + 165,79 seconds

Time²  - 0,4737 Time² = 165,79 seconds

0,5263 Time² = 165,79 seconds

Time² = 165,79 seconds/0,5263 =315,01 seconds

Distance=1,9 x 315,01 seconds = 598,51 m

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Answer:

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Step-by-step explanation:

Surface area of a cube,A = 6s²

Where,

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Cube X has a side length of 1,

Surface area of cube, X = 6s²

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How much greater is the surface area of cube Y than cube X?

Surface area of Cube Y is 3 times greater than the surface area of cube X

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Ron has a bag containing 3 green pears and 4 red pears he randomly selects a lead then randomly selects another pear without rep
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First pear            Second pear      probability

green (3/7)---------green (2/6)         (3/7)*(2/6) = 1/7
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red (4/7)-------------green (3/6)        (4/7)*(3/6) = 2/7
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3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

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Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
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