A lock has a code of 4 numbers between 1 and 20. If no numbers in the

Can someone please help me. I need help on the 5 one

soldi70 [24.7K]

Answer:

The second option

Step-by-step explanation:

On the right hand side you have all the multiples of 4, on the left hand side you have 5

The multiples of four are ...,-8,-4,0,4,8,....

so the simbol that means that 5 is not an element of the multiples of four is the second option.

4 0

3 years ago

Please Help! 30 points! Real answers pls The box plot below shows the total amount of time, in minutes, the students of a class

padilas [110]

Hii!

So I did this and I have A!

A: You cannot get the mean from the graph but you CAN get the third quartile!

B: To find the interquartile range (IQR) we subtract the third and first quartiles:

60-35 = 25

C: An outlier would be much larger than the rest of the data or much smaller than the rest of the data. An outlier would make the "whisker" portion longer and could potentially slightly shift the box.

6 0

3 years ago

Read 2 more answers

What is the circumference of this circle d=12 cm

Gnoma [55]

For circumference it’s 12x3.14= 37.68

6 0

3 years ago

Read 2 more answers

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for <em>h: </em>

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for <em>h</em>.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

7 0

3 years ago

Madison is writing a report on the human body for science class. She wants to use numbers in exponent form for her report. She i

kkurt [141]

75 x 104 . . . ..........................

3 0

3 years ago