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The popular candy Skittles comes in 5 colors. According to the Skittles website, the 5 colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population. Suppose that we purchase a small bag of Skittles. Assume this size bag always has 40 candies. In this particular bag 10 are green. What is the probability that a randomly selected bag of this size has 10 or more green candies

Answers

Answer 1

Answer:

Answer:

27.76% probability that a randomly selected bag of this size has 10 or more green candies

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$√(V(X)) = √(np(1-p))$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = √(V(X))$ .

In this problem, we have that:

$n = 40, p = 0.2$

$\mu = E(X) = np = 40*0.2 = 8$

$\sigma = √(V(X)) = √(np(1-p)) = √(40*0.2*0.8) = 2.53$

What is the probability that a randomly selected bag of this size has 10 or more green candies

Using continuity correction, this is $P(X \geq 10 - 0.5) = P(X \geq 9.5)$ , which is 1 subtracted by the pvalue of Z when X = 9.5. So

$Z = (X - \mu)/(\sigma)$

$Z = (9.5 - 8)/(2.53)$

$Z = 0.59$

$Z = 0.59$ has a pvalue of 0.7224

1 - 0.7224 = 0.2776

27.76% probability that a randomly selected bag of this size has 10 or more green candies

Answer 2

Answer:

Answer:

$P(x\geq 10)=0.2682$

Step-by-step explanation:

The number x of green candies in a bag of 40 candies follows a binomial distribution, because we have:

n identical and independent events: 40 candies
a probability p of success and (1-p) of fail: a probability of 0.2 to get a green candie and 0.8 to doesn't get a green candie.

So, the probability that in a bag of 40 candies, x are green is calculated as:

$P(x)=(n!)/(x!(n-x)!)*p^(x)*(1-p)^(n-x)$

Replacing, n by 40 and p by 0.2, we get:

$P(x)=(40!)/(x!(40-x)!)*0.2^(x)*(1-0.2)^(40-x)$

So, the probability that a randomly selected bag of this size has 10 or more green candies is equal to:

$P(x\geq 10)=P(10)+P(11)+...+P(40)\nP(x\geq 10)=1-P(x<10)$

Where $P(x<10)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)$

So, we can calculated P(0) and P(1) as:

$P(0)=(40!)/(0!(40-0)!)*0.2^(0)*(1-0.2)^(40-0)=0.00013\nP(1)=(40!)/(1!(40-1)!)*0.2^(1)*(1-0.2)^(40-1)=0.00133$

At the same way, we can calculated P(2), P(3), P(4), P(5), P(6), P(7), P(8) and P(9) and get that P(x<10) is equal to:

$P(x<10)=0.7318$

Finally, the probability $P(x\geq 10)$ that a randomly selected bag of this size has 10 or more green candies is:

$P(x\geq 10)=1-P(x<10)\nP(x\geq 10)=1-0.7318\nP(x\geq 10)=0.2682$

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Answers

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Never want to give you up

Step-by-step explanation:

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To make lemonade you can mix 4 teaspoons of lemonade powder with 16 ounces of water. What is the ratio of powder towater?
4:32
32:8
24:64
32:128

Answers

Answer:

32:128

Step-by-step explanation:

divide all of it by 2, you get 16:64. Again, 8:32. Again, 4:16

Peter baked 64 loaves of bread in three days. How many loaves did he bake each day, if he baked 3 more loaves on the second day than on the first day, and 4 more loaves on the third day than on the first day?

Answers

Answer:

Step-by-step explanation:

1st day: x

2nd day: x+3

3rd day: x+4

Equation: x+(x+3)+(x+4)=64

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I hope you found this answer helpful!!!!!!(sorry if instructions aren't clear)

Estimate 2/3% of $141

Answers

Answer:

0.94 cents

Step-by-step explanation:

If you multiply 0.67 with 141 you will see that you will end up with the same answer as above.

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1. Consider the following hypotheses:H1 : ∃x (p(x) ∧ q(x)) H2 : ∀x (q(x) → r(x))
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (p(x) ∧ r(x))
Clearly label the inference rules used at every step of your proof.

2. Consider the following hypotheses:
H1 : ∀x (¬C(x) → ¬A(x)) H2 : ∀x (A(x) → ∀y B(y)) H3 : ∃x A(x)
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (B(x) ∧ C(x))
Clearly label the inference rules used at every step of your proof.

3. Consider the following predicate quantified formula:
∃x ∀y (P (x, y) ↔ ¬P (y, y))
Prove the unsatisfiability of this formula using rules of inference.

Answers

Answer:

See deductions below

Step-by-step explanation:

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f) ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

Which of the following is a like radical to 3 x StartRoot 5 EndRoot?x (RootIndex 3 StartRoot 5 EndRoot)
StartRoot 5 y EndRoot
3 (RootIndex 3 StartRoot 5 x EndRoot)
y StartRoot 5 EndRoot

Answers

Answer:

Step-by-step explanation:

Final answer:

The like radical to the expression 3√5 is y√5, as both expressions have the square root index and the same radicand, which is 5.

Explanation:

The student is asking which radical expression is like the radical 3√5. Like radicals have the same index and radicand. The index is the degree of the root, and the radicand is the number under the radical sign. The expression 3√5 means 3 times the square root of 5, or in exponential form, 3 × 51/2. The like radical for 3√5 would also need to have a square root (index of 2) and the same radicand (5). Therefore, the like radical to 3√5 from the options provided would be y√5 because it has the same index (2) and radicand (5), only with a different coefficient (y instead of 3).

Additionally, expressing radicals as fractional exponents helps to identify like radicals. For example, using the property x² = √x we can understand that if we have the same base and exponent, we can consider the expressions to be like radicals. Hence, in this case, since both 3 and y are just coefficients, the root parts √5 are the same, making them like radicals.

Learn more about Like Radicals here:

brainly.com/question/30277929

#SPJ3

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