All categories

3 years ago

Find the median for the given set of data.5/8,3/4,1 1/8,1 1/4

Mathematics

1 answer:

Basile [38]3 years ago

8 0

Answer:

Step-by-step explanation: 15/16

I added the numbers and got 30/8

30 divided by the number of numbers (4) is 7.5

Since we are not dealing in eighths, we extrapolate that 7.5 eights = 15 sixteens, or

15/16

You might be interested in

Fully simplify. -15yº (5x²y)

miss Akunina [59]

Answer: (−15y)(5x2y)

=−75x2y2

7 0

4 years ago

Help please I really can’t fail this exam

shtirl [24]

Answer:E'F' and EF are equal in length.

7 0

4 years ago

Directed line segment overline FG has endpoints F(4, 9) and G(- 8, - 9) .On the coordinate plane below , plot a point that parti

Thepotemich [5.8K]

Answer:

The coordinates of the point that makes the division in the given ratio is (0,3)

Step-by-step explanation:

Here, we want to find the point on the line segment that divides the line segment in the ratio 1:2

We simply use the internal division formula

That would be;

(x,y) = (nx1 + mx2)/(m + n) , (ny1 + my2)/(m + n)

m = 1 and n = 2

(x1,y1) = (4,9)

(x2,y2) = (-8,-9)

Substituting these values into the formula, we have;

2(4) + 1(-8)/(1 + 2) , 2(9) + 1(-9)/(2 + 1)

= (8-8)/3 , (18-9)/3

= (0,3)

5 0

3 years ago

Which of the following is true of the commutative property under subtraction

Yuri [45]

The answer is A I believe it is

3 0

4 years ago

Read 2 more answers

A steel safe with mass 2200 kg falls onto concrete. Just

Virty [35]

The kinetic energy of the safe increases the force exerted by the concrete

to several times the weight of the safe.

The magnitude of the force exerted on the safe by the concrete on the is approximately $\underline{29.\overline 3 \, \mathrm{MN}}$

The concrete exerts a force that is approximately 1,359.16 times the weight of the safe.

Reasons:

First part

The mass of the steel safe, m = 2,200 kg

Velocity of the safe just before it hits the concrete, v = 40 m/s

The amount by which the safe was compressed, d = 0.06 m

The kinetic energy, K.E., of the safe just before it hits the round is therefore;

$\displaystyle K.E. = \mathbf{\frac{1}{2} \cdot m \cdot v^2}$

$\displaystyle K.E._{safe} = \frac{1}{2} \times 2,200 \times 40^2 = 1,760,000 \ Joules$

Work done by concrete, W = Force, F × Distance, d

$\displaystyle Force, \, F = \mathbf{\frac{Work, \, W}{Distance, \, d}}$

By the law of conservation of energy, we have;

The work done by the concrete, W = The kinetic energy, K.E. given by the safe

W = K.E. = 1,760,000 J

The effect of the work = The change in the height of the safe

Therefore;

The distance, d, over which the force of the concrete is exerted = The change in the height of the safe = 0.06 m

d = 0.06 m

Therefore;

$\displaystyle The \ force \ of \ the \ concrete, \, F = \frac{1,760,000\, J}{0.06 \, m} = 29,333,333. \overline 3 \, N = 29.\overline 3 \ MN$

The force of the concrete on the safe = $\underline{29.\overline 3 \ MN}$

Second part:

The gravitational force of the Earth on the safe, W = The weight of the safe

W = Mass, m × Acceleration due to gravity, g

W = 2,200 kg × 9.81 m/s² ≈ 21,582 N

The ratio of the force exerted by the concrete to the weight of the safe is found as follows;

$\displaystyle Ratio \ of \ forces = \frac{29.\overline 3 \times 10^6 \, N}{21,582 \, N} = \frac{4,000,000}{2,943} \approx \mathbf{1359.16}$

The force exerted by the concrete is approximately 1,359.16 times the weight of the safe.

Learn more here:

brainly.com/question/21060171

5 0

3 years ago