The volume of the box below is 1.5 cubic inches. Which of the following are possible dimensions of the box.

Jessica packed 1-inch cubes into a box with a volume of 144 cubic inches how many layers of 1-inc cubes did Jessica pack

alexdok [17]

Maybe 3 or 6 or 8, the formula for the volume of a rectangular prism = a * b * c. A, B, C are the length, width, and height. and 3*6*8=144

7 0

3 years ago

Maritza remembers her pin because it is the product of consecutive prime numbers that is between 1000 and 1500. What is her pin

faltersainse [42]

Well, i am not sure if i am right, but i tried.

The prime numbers have to be consecutive, (meaning in order from least to greatest), so you find 2 prime numbers on a list, and the only two are 31 and 37 so multiply those and you should get 1147

8 0

3 years ago

which data set has a variation, or mean absolute deviation, similar to the data set given on the dot plot

inessss [21]

Do you know the answer I have the same question

6 0

2 years ago

Read 2 more answers

Evaluate 20/6 - 2/3. 2 1/6, 2 2/3, 3 1/3, 3 2/3

N76 [4]

Answer:

Step-by-step explanation:

20/6 - 4/6= 16/6= 2 4/6= 2 2/3

7 0

2 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

3 years ago