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3 years ago

12

There are 5 balls, 3 red, and 2 black. What is the probability that a random ordering of the 5 balls does not have the 2 black b

alls next to each other

1 answer:

Molodets [167]3 years ago

8 0

Answer:

there are 6 different ways they will not be next to each other

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masya89 [10]

Answer:

Part A = the middle one

Part B = 4 days

Step-by-step explanation:

1,300 / 4 = 325

325 x 4 = 1,300

8 0

3 years ago

Read 2 more answers

15. In the diagram AB is parallel to CD.<br> Calculate the value of a.<br> NOT TO<br> SCALE

VikaD [51]

Same-side interior angles, which are shown here, are supplementary, which means that they add up to 180 degrees.

5a + a = 180

6a = 180

a = 30 degrees

Hope this helps! :)

5 0

3 years ago

Mason used 1/6 of a gram of honey to make 1/12 of a pan of pudding. How many grams are needed to make a full pan of pudding?

NARA [144]

The answer would be two grams of honey, because 1 gram of honey would make 1 half pan (because 6 is half of 12) So therefore, two grams of honey would make one pan of pudding.

7 0

3 years ago

1. The Sharks Aquatic Club recently held a fundraiser to raise money for a local charity. The swimmers received money for each l

liq [111]

Answer:

Rita = 300

John = 900

Rodell = 925

Step-by-step explanation:

rita = x (let

john = 3x ( john swims three times more than rita

rodell = 25 + 3x ( rodell swims 25 more than john

x + 3x + 25 + 3x = 2125

or, 7x + 25 = 2125

or, 7x = 2125 - 25

or, x = 2100/7

x = 300

hence,

x = 300

3x = 3 × 300

= 900

25 + 3x = 25 + 900

= 925

8 0

3 years ago

21. Who is closer to Cameron? Explain.

pickupchik [31]

Problem 21

<h3>Answer: Jamie is closer</h3>

-----------------------

Explanation:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

$A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\$

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

$J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\$

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer: Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)
B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

AB
JB
CB

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

$A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\$

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

AB = 29.1547595
JB = 41.2310563
CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0

2 years ago