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blsea [12.9K]
2 years ago
6

On a December day, the probability of snow is 0.30. The probability of a "cold" day is 0.50. The probability of snow and "cold"

weather is 0.15. Are snow and "cold" weather independent events?a only if given that it snowedb Noc Yesd only when they are also mutually exclusive
Mathematics
1 answer:
Margarita [4]2 years ago
8 0

Answer:

Yes, snow and cold weather are independent.

Step-by-step explanation:

We are given the following in the question:

C: Cold weather

S: Snow

P(C) = 0.50

P(S) =0.30

P(S\cap C) = 0.15

We have to check whether snow and cold whether are independent events.

If the events A and B are independent then,

p(A\cap B) = P(A)\times P(B)

Checking,

p(S\cap C) = P(S)\times P(C)\\0.15 = 0.30\times 0.50

Thus, the two events are independent.

For mutually exclusive events

P(A\cap B) = 0

Thus, the given events are not mutually exclusive.

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The population P(t) of a bacteria culture is given by P(t)=−1,500t2+48,000t+18,000, where t is the time in hours after the cultu
Anna [14]

Answer:

this is a tough question

6 0
2 years ago
The combined average weight of an okapi and a llama is 450450450 kilograms. The average weight of 333 llamas is 190190190 kilogr
Kisachek [45]
I'm going to assume that you meant 450kg for the combined weight, 190kg more and 3 Llamas. I'm pretty sure Llamas and Okapis don't weigh 450450450kg (that's 993,073,252 pounds). :)

x= Okapi weight
y= Llama weight

EQUATIONS:
There are 2 equations to be written:

1) 450kg is equal to the weight of one Okapi and one Llama

450kg= x + y

2) The weight of 3 llamas is equal to the weight of one Okapi plus 190kg.

3y=190kg + x


STEP 1:
Solve for one variable in one equation and substitute the answer in the other equation.

450kg= x + y
Subtract y from both sides
450-y =x


STEP 2:
Substitute (450-y) in second equation in place of x to solve for y.

3y=190kg + x
3y=190 + (450-y)
3y=640 -y
add y to both sides

4y=640
divide both sides by 4

y=160kg Llama weight


STEP 3:
Substitute 160kg in either equation to solve for x.

3y=190kg + x
3(160)=190 + x
480=190 + x
Subtract both sides by 190

290= x
x= 290kg Okapi weight


CHECK:
3y=190kg + x
3(160)=190 + 290
480=480

Hope this helps! :)
8 0
2 years ago
The marching band walks 1/15 miles in 1.5 minutes. At this rate, how many minutes will it take to complete 1 mile? Record your a
ipn [44]

Answer:

It will take 22.5 minutes to complete 1 mile

Step-by-step explanation:

From the question,

The marching band walks 1/15 miles in 1.5 minutes

First we will determine the miles cover in 1 minute

If the marching band walks 1/15 miles in 1.5 minutes

then, they will cover x miles in 1 minute

x = \frac{1/15 * 1}{1.5}\\

Then,

x = \frac{1/15 }{1.5}

x = 2/45 miles

∴ 2/45 miles are covered in 1 minute

Now,

If 2/45 miles are covered in 1 minute

Then, 1 mile will be covered in y minute

y = \frac{1 * 1}{2/45}

y =\frac{1}{2/45} \\y = \frac{45}{2}\\

∴ y = 22.5 minutes

Hence, it will take 22.5 minutes to complete 1 mile

6 0
2 years ago
The equation 1.5r+15=2.25r1.5r+15=2.25r represents the number rr of movies you must rent to spend the same amount at each movie
Lelu [443]
1.5r+15=2.25r
Combine like terms: 1.5r+15-1.5r=2.25r-1.5r
15=0.75r
Get the unknown alone: 15/.75=.75/.75r
20=r or r=20 :)

7 0
2 years ago
hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a
nalin [4]

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

6 0
2 years ago
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