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Zigmanuir [339]

2 years ago

11

What I'd the value of x

1 answer:

satela [25.4K]2 years ago

5 0

Answer:

Can you please give us figure so that i can help you

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1. A student is taking a multiple-choice exam in which each question has four choices. Assume that the student has no knowledge

LenKa [72]

Answer:

a) 0.001 = 0.1% probability that she will get five questions correct.

b) 0.0156 = 1.56% probability that she will get at least four questions correct.

c) 0.2373 = 23.73% probability that she will get no questions correct.

d) 0.8965 = 89.65% probability that she will get no more than two questions correct.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either she gets it correct, or she does not. The probability of getting a question correct is independent of any other question, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

There are five multiple choice questions on the exam.

This means that $n = 5$

She has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball for each question and replaces the ball in the box.

This means that $p = \frac{1}{4} = 0.25$

a. Five questions correct?

This is $P(X = 5)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

0.001 = 0.1% probability that she will get five questions correct.

b. At least four questions correct?

This is:

$P(X \geq 4) = P(X = 4) + P(X = 5)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0146 + 0.001 = 0.0156$

0.0156 = 1.56% probability that she will get at least four questions correct.

c. No questions correct?

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

0.2373 = 23.73% probability that she will get no questions correct.

d. No more than two questions correct?

This is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

$P(X = 1) = C_{5,0}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2373 + 0.3955 + 0.2637 = 0.8965$

0.8965 = 89.65% probability that she will get no more than two questions correct.

3 0

3 years ago

PLZ ANSWER ASAP

Nutka1998 [239]

Answer:

C.)-18 $\sqrt{7}$ Is your answer!

Step-by-step explanation:

Please mark brainliest if this helped! :D

3 0

3 years ago

Read 2 more answers

Solve the original equation 2+1/b+2=3b/b+2 by solving the proportion. The solutions are

Damm [24]

Answer:

b=1

Step-by-step explanation:

2b+1+2b=3b+2b

4b+1=5b(Simplify both sides of the equation)

4b+1−5b=5b−5b(Subtract 5b from both sides)

−b+1=0

−b+1−1=0−1(Subtract 1 from both sides)

−b=−1

−b

−1

=

−1

−1

(Divide both sides by -1)

b=1

Check answers. (Plug them in to make sure they work.)

b=1(Works in original equation)

6 0

3 years ago

Read 2 more answers

Please help me on number 1

dolphi86 [110]

The answer is C. This is because -4 is less than 0, so there can't be any counters if the value is less than 0.

6 0

3 years ago

The high-school football team scored 19 points on touchdowns (worth 7 points if you count the extra point) and field goals (wort

Kazeer [188]

If they scored 5 times, they got 1 touchdown and 4 field goals.

Equation:

7 + 3 + 3 + 3 + 3 = 19.

5 0

3 years ago