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3 years ago

If a card is picked at random from a standard 52-card deck, what is the probability of getting a spade or a King?

1 answer:

8 0

There are 13 spades in a deck, so the probability is 1/4. There are 4 King’s in the deck, but a king can also be a spade, so what would be 1/13 is actually 3/52. So add the probabilities together, and you get 13/52 + 3/52 = 16/52, or 4/13

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Write 235 using powers of 10

vaieri [72.5K]

Answer:

(10^2)*2, 10*3, 10/2

Step-by-step explanation:

Thats my best guess to your very vague question

8 0

3 years ago

A regular 18-sided polygon is rotated with the center of rotation at its center. What is the smallest degree of rotation needed

dem82 [27]

It will be easier to explain how to solve this question in a square. If you rotate 4-sided square from the centre, you need to rotate it 90 degrees. The formula for this would be: 360 ° /the side count. In a square, it would be 360° /4= 90°.

In 18-sided polygon, the calculation would be: 360°/ 18 side= 20°

4 0

3 years ago

Read 2 more answers

A certain three-cylinder combination lock has 60 numbers on it. To open it, you turn to a number on the first cylinder, then

statuscvo [17]

Answer:

(a) The number of different lock combinations is 216,000.

(b) The probability of getting the correct combination in the first try is $\frac{1}{216000}$ .

Step-by-step explanation:

The lock has three-cylinder combinations with 60 numbers on each cylinder.

The procedure of the opening lock is to turn to a number on the first cylinder, then to a second number on the second cylinder, and then to a third number on the third cylinder.

The numbers can be repeated.

(a)

There are three cylinder combinations on the lock, each with 60 numbers.

It is provided that repetitions are allowed.

Then each cylinder can take any of the 60 numbers.

So there are 60 options for the first cylinder.

There are 60 options for the second cylinder.

And there are 60 options for the third cylinder.

So the total number of possible combinations is:

Total number of combinations = 60 × 60 × 60 = 216000.

Thus, the number of different lock combinations is 216,000.

(b)

The events of getting a correct combinations of the three-cylinder combination lock implies that all the three cylinder are set at the correct numbers.

Each cylinder has 60 numbers.

This implies that there are 60 possible ways to get a correct number for the first cylinder.

The probability of getting the correct number for the first cylinder is:

P (Number on 1st cylinder is correct) = $\frac{1}{60}$ .

Similarly for the second cylinder the probability of getting the correct number is:

P (Number on 2nd cylinder is correct) = $\frac{1}{60}$ .

And similarly for the third cylinder the probability of getting the correct number is:

P (Number on 3rd cylinder is correct) = $\frac{1}{60}$ .

So the probability of getting the correct combination in the first try is:

P (Correct combination in the 1st try) = $\frac{1}{60}\times \frac{1}{60}\times \frac{1}{60}=\frac{1}{216000}$ .

Thus, the probability of getting the correct combination in the first try is $\frac{1}{216000}$ .

7 0

3 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

4 0

3 years ago

Hi, umm pls help 1/6(x-5)=1/2(x+6), solve for x And can you please explain

Mumz [18]

Answer:

By solving the equation $\frac{1}{6}(x-5)=\frac{1}{2}(x+6)$ we found that $x=\frac{23}{2} \text { or }=11 \frac{1}{2}$

Explanation:

Given equation,

$\frac{1}{6}(x-5)=\frac{1}{2}(x+6)$ to find x

Step: 1 Cross multiply the denominators 2(x-5)=6(x+6)

Step: 2 Open brackets and simplify the term 2x-10=6x+36

Step: 3 Bring x terms on one side and simplify the equation the equation obtained is following,

2x-6x=36+10

-4x=46

$-x=\frac{46}{4}=\frac{23}{2}=11 \frac{1}{2}$

Therefore, we solved x value from the given equation $\frac{1}{6}(x-5)=\frac{1}{2}(x+6)$ as $x=\frac{23}{2} \text { or }=11 \frac{1}{2}$ .

4 0

3 years ago