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3 years ago

it took team infinite dimensions 2 hours to hike 4 2/3 miles what is their average speed in miles per hour

1 answer:

5 0

To find their mph you would have to divide by 2 to find out how much they got in 1 hour instead of 2. You would have to do 4 2/3 / 2. An easier way to do this is just doing 14/3 x 1/2 which is the same thing except making 4 2/3 a fraction and changing divided by 2 to multiply 1/2. The answer you would get is 14/6, which is 2 1/3 miles per hour. To do this quickly you could just look at the question and divide 4 by 2 to get 2 and then divide 2/3 by 2 to get 1/3 and then add them up.
Hope this helps!

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Q and R are not mutually exclusive events. If P(Q)=5/8 and P(R)=1/8 and P(Q and R)=1/32,Find P(q or R)

artcher [175]

The solution for this problem is just adding and subtracting fractions. We all know how to add and subtract function because our teacher since high school teach us how to. P(Q or R) = P(Q) + P(R) - P(Q and R) = 5/8 + 1/8 - 1/32 = 23/32

The answer to this question is 23/32.

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3 years ago

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Find the equation of the perpendicular bisector of a line segment whose end points are (-3,9) and (-1,5)

timofeeve [1]

First find the equation of y.

$y=mx+n=\frac{\Delta y}{\Delta x}x+n$

Find the slope m.

$m=\frac{5-9}{-1-3}=\frac{-4}{-4}=1$ .

Pick one point, I'll pick (-3, 9). Insert coordinates in equation then compute n.

$9=1(-3)+n\implies n=12$ .

The equation of a line y is:

$y=x+12$ .

The perpendicular line $y_{\perp}$ is same like the normal line except its slope m becomes:

$k=-\frac{1}{m}=-1$ .

The equation of a perpendicular bisector is thus:

$y_{\perp}=-x+12$ .

Hope this helps.

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4 years ago

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100000000000000000000+10000000000000000

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1.0001e+20

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si tienes una mesa de 25m de largo por 10 de ancho ¿ en cuanto varíaran cada una de sus dimensiones para que sin alterar su supe

Murrr4er [49]

responder: 8,75

Cada dimensión será de 8,75 porque si la longitud es de 25 m de largo y el ancho es 10, si los sumamos y obtendremos 35. Así que 35 dividido entre 4 es 8,75

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4 years ago

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