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sukhopar [10]
1 year ago
5

5x5x5x5x5x6x7x8x9x1x2x3

Mathematics
1 answer:
MrRa [10]1 year ago
5 0
Well by the powers to each of these numbers the only you have to multiply the answer by the next answer. This one took some time but i got 56,700,00 i also calculated the answer to see was it correct. It is. Hope this helps!
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Eighteen 2.5 gallon buckets are needed to fill a cistern with water. Find the constant of variation. Please help! Thank you!
Allushta [10]

Given:

Eighteen 2.5 gallon buckets are needed to fill a cistern with water.

To find:

The constant of variation.

Solution:

If y is directly proportional to x, then

y\propto x

y=kx

Where, k is constant of variation.

In the given problem, water in cistern (w) is directly proportional to number of buckets (n).

w\propto n

w=2.5n        (Capacity of each bucket is 2.5 gallons)

Therefore, the constant of variation is 2.5.

6 0
1 year ago
Read 2 more answers
1.17 A study of the effects of smoking on sleep patterns is conducted. The measure observed is the time, in minutes, that it tak
bearhunter [10]

Answer:

a.

\bar X_F=43.7

\bar X_{NF}=30.32

b.

S_F=16.9278

S_{NF}=7.12783

c.

Attached file

d.

Apparently the practice of smoking reduces the ability to fall asleep, demanding much more time in individuals who smoke, than in those who do not smoke.

Step-by-step explanation:

a, b) For the group of smoking individuals, the average time it takes to fall asleep and the standard deviation of those times is:

\bar X_F={\frac{1}{n} \sum_{i=1}^n x_i = 43.7

S_F=\sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})^2}=16.9278

a, b) For the group of non-smoking individuals, the average time it takes to fall asleep and the standard deviation of those times is:

\bar X_{NF}={\frac{1}{n} \sum_{i=1}^n x_i = 30.32

S_{NF}=\sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})^2}=7.12783

c. In the attached file you can see the diagram of points for the times, in the smoking and non-smoking groups.

d. Apparently the practice of smoking reduces the ability to fall asleep, demanding much more time in individuals who smoke, than in those who do not smoke.

Download pdf
6 0
2 years ago
Sandy evaluated the expression below. (negative 2) cubed (6 minus 3) minus 5 (2 + 3) = (negative 2) cubed (3) minus 5 (5) = 8 (3
nasty-shy [4]

Answer:

should be - 8

Step-by-step explanation:

-2*-2=4 4*-2=-8

4 0
2 years ago
Read 2 more answers
A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each r
bulgar [2K]

Answer:

The answer to your question is 8 h

Step-by-step explanation:

Data

length of the ladder = 200 cm

distance between each rung = 20 cm

rate = 10 cm/h

fifth rung = ?

Process

1.- Calculate the total distance the tide must rise

distance = 20 cm x 4

              = 80 cm              because the first rung touches the water

2.- Calculate the time

rate = distance / time

-Solve for time

time = distance / rate

-Substitution

time = 80 cm / 10cm/h

-result

time = 8 h

3 0
1 year ago
a ball is thrown with a slingshot at a velocity of 110ft/sec at an angle of 20 degrees above the ground from a height of 4.5 ft.
satela [25.4K]

Answer:

t=2.47\ s  

Step-by-step explanation:

The equation that models the height of the ball in feet as a function of time is

h(t) = h_0 + s_0t -16t ^ 2

Where h_0 is the initial height, s_0 is the initial velocity and t is the time in seconds.

We know that the initial height is:

h_0 = 4.5\ ft

The initial speed is:

s_0 = 110sin(20\°)\\\\s_0 = 37.62\ ft/s

So the equation is:

h (t) = 4.5 + 37.62t -16t ^ 2

The ball hits the ground when when h(t) = 0

So

4.5 + 37.62t -16t ^ 2 = 0

We use the quadratic formula to solve the equation for t

For a quadratic equation of the form

at^2 +bt + c

The quadratic formula is:

t=\frac{-b\±\sqrt{b^2 -4ac}}{2a}

In this case

a= -16\\\\b=37.62\\\\c=4.5

Therefore

t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}

t_1=-0.114\ s\\\\t_2=2.47\ s  

We take the positive solution.

Finally the ball takes 2.47 seconds to touch the ground

4 0
2 years ago
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