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expeople1 [14]
2 years ago
6

Please HELP ME WITH THIS TY!

Mathematics
1 answer:
LenKa [72]2 years ago
7 0
X=117 degrees
Y = 63 degrees
Z = 117 degrees
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A bus makes a stop at 2:30, letting off 10 people and letting on 2. The bus makes another stop ten minutes later to let off 3 mo
maksim [4K]

Answer:

Step-by-step explanation:

B

5 0
3 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

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

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

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
A box contains 2 red marbles, 3 green marbles, and 3 blue marbles. if we choose a marble, then another marble without putting th
Bad White [126]
There are 7 marbles, including 3 green ones, so at the start, there is a 3/7 chance of getting a green marble.
Assuming you did get a green one, there are now 6 marbles left, with 2 blue marbles, so there is a 2/6 chance of taking a blue marble.
Given that both have to happen, you must multiply each probability, hence the total probability is 3/7 x 2/6, or a 1/7 chance.
Hope this helped
7 0
3 years ago
Read 2 more answers
Find the length of the third side if necessary write in simplest radical form ​
KATRIN_1 [288]

Answer:

6²-3²

36-9

<h2>=√27</h2>

Step-by-step explanation:

might be Wright answer

according to pythogorus therom

5 0
2 years ago
How can you write the expression with rationalized denominator? sqrt3 - sqrt6 / sqrt 3 + sqrt 6
harina [27]
Answer:  2√2 - 3

Explanation:


The expession written properly is:

\frac{ \sqrt{3}- \sqrt{6}  }{ \sqrt{3}+ \sqrt{6} }

To rationalize that kind of expressions, this is to eliminate the radicals on the denominator you use conjugate rationalization.

That is, you have to multiply both numerator and denominator times the conjugate of the denominator.

The conjugate of √3+√6 is √3 - √6, so let's do it:

\frac{ \sqrt{3} - \sqrt{6} }{ \sqrt{3} + \sqrt{6} } . \frac{ \sqrt{3}- \sqrt{6}  }{ \sqrt{3}- \sqrt{6} }

To help you with the solution of that expression, I will show each part.

1) Numerator: (√3 - √6) . (√3 - √6) = (√3 - √6)^2 = (√3)^2 - 2√3√6 + (√6)^2 =

= 3 - 2√18 + 6 = 9 - 6√2.

2) Denominator: (√3 + √6).(√3 - √6) = (√3)^2 - (√6)^2 = 3 - 6 = - 3

3) Then the resulting expression is:

 9 - 6√2
-----------
 -3

Which can be further simplified, dividing by - 3

    -3 + 2√2

Answer: 2√2 - 3

7 0
3 years ago
Read 2 more answers
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