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3 years ago

Which equation is another way to show 2/3 A. 1/2 ÷ 3 = _ B. 1/3 ÷ 2 = _ C. 2 ÷ 1/3 = _ D. 2 ÷ 3 = _ E. 3 ÷ 2 = ___

Mathematics

1 answer:

gulaghasi [49]3 years ago

3 0

Answer:

Step-by-step explanation:

A. 1/2 ÷ 3 = 1/6

B. 1/3 ÷ 2 = 1/6

C. 2 ÷ 1/3 = 6/1

D. 2 ÷ 3 = 2/3

E. 3 ÷ 2 = 3/2

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What is the median of this data set?

bazaltina [42]

Answer:

Step-by-step explanation:

Because the median is the middle number when the data set is in order least to greatest so you already have it in least to greatest so thank you for helping me.

and for extra just in case.

Mean=16.222

the mean= the average of the data set

Mode is 13,14

The mode is the number that appears the most in a data set and 13 and 14 appear twice.

Your range is=13

Range is the highest number subtracting the lowest number so 24 -11 which =13

Minimum is 11 because the minimum is the lowest number in data set..

Maximum is 24

Maximum is the highest number in the data set.

I hope this helps have a awesome day:)

7 0

2 years ago

Read 2 more answers

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago

In Texas, the distance from Austin to Brownwood is 141 miles. It is 171 miles from Brownwood to Corsicana and 157 miles from Cor

Alecsey [184]

Answer: acute angles

Step-by-step explanation:

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3 years ago

Can absolute value be negative

emmasim [6.3K]

No, only the value of what you are finding the absolute value of can be a negative. Examples: |-2|=2 |2|=2

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3 years ago

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horrorfan [7]

The first one is 20. The second one is 64 points

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3 years ago

Which equation is another way to show 2/3 A. 1/2 ÷ 3 = ___ B. 1/3 ÷ 2 = ___ C. 2 ÷ 1/3 = ___ D. 2 ÷ 3 = ___ E. 3 ÷ 2 = ___​

Which equation is another way to show 2/3 A. 1/2 ÷ 3 = _ B. 1/3 ÷ 2 = _ C. 2 ÷ 1/3 = _ D. 2 ÷ 3 = _ E. 3 ÷ 2 = ___