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PolarNik [594]
3 years ago
15

I need 20 characters to post this so ignore

Mathematics
1 answer:
Natali5045456 [20]3 years ago
6 0
KSKDWKDKSODMEMFKSCK HHHHhhhHhHhHhHhhsenfjsockwmf
You might be interested in
Cos(0.5)+cos(0.5+2(3.14159265))+cos(0.5+4(3.14159265))
aleksandr82 [10.1K]

Answer:

2.63274768567

Step-by-step explanation:

First, note that 3.1415... is p, so 2(3.1415...) is 2pi and 4(3.1415...) is 4pi.

2pi and 4pi are full rotations, which will lead to the same cosine (i.e. cos(x) = cos (x+2pi) = cos(x+4pi) = ... = cos (x+2k*pi)).

So, the expression equals cos0.5 + cos 0.5 + cos0.5 = 3cos0.5 = 3(0.87758256189) = 2.63274768567

I hope this helps! :)

7 0
3 years ago
A jar has 6 red marbles and 4 blue marbles. If 2 marbles are chosen, one at a time without replacement, what is the probability
Juliette [100K]

1/3 Probability both are red

red picked first:

6/10

red picked second:

5/9 (9 as first red marble not replaced so 5 red marbles remaining and 4 blue marbles still)

6/10 * 5/10 = 1/3

Hope that helps!!

7 0
3 years ago
At a video arcade action video games usually cost $12.50. This week they are on sale for 30% off. What is the sale price of the
Juliette [100K]

Answer: 8.75


Step-by-step explanation:

12.50 x 30% or .03

= 3.75

3.75 - 12.50

= 8.75

5 0
3 years ago
Read 2 more answers
A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra
VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that N = 6

4 blue, which means that k = 4

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that N = 6

2 yellow, which means that k = 2

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that P(A) = 0.4 + 0.0667 = 0.4667

Probability of both being the same color and blue:

40% probability that both are blue, which means that P(A \cap B) = 0.4

Desired probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0
3 years ago
How do you solve?3(t+4)-2(2t+3)=-4
shepuryov [24]
3t+12-4t-6=-4
-t=-4+6-12
-t=-10
t=10
That's your answer.
8 0
3 years ago
Read 2 more answers
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