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Kitty [74]
3 years ago
15

Order of operations A set of three scores consists of the values 6, 3, and 2. 1. Σ2X – 2 =______.a. 35.b. 33.c. 22.d. 6.2. Σ(2X)

² =______.a. 248.b. 124.c. 61.d. 39.
Mathematics
1 answer:
Nostrana [21]3 years ago
4 0

Answer:

Step-by-step explanation:

Given the order of operations A set of three scores to consists of the values 6, 3, and 2, we are to evaluate the following

1) Σ2X – 2

= 2(6)+2(3)+2(2) – 2

= 12+6+4–2

= 22–2

= 20

2) Σ(2X)² = (2×6)²+(2×3)²+(2×2)²

Σ(2X)² = 12²+6²+4²

Σ(2X)² = 144+36+16

Σ(2X)² = 144+52

Σ(2X)² = 196

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The probability of winning a lottery by selecting the correct six integers, are

\begin{aligned}&(a) 1.68 \times 10^{-6} \\&(b) 5.13 \times 10^{-7} \\& (c) 1.91 \times 10^{-7} \\&(d)8.15 \times 10^{-8}\end{aligned}

<h3>What is binomial distribution?</h3>

The binomial distribution is a type of probability distribution that expresses the probability that, given a certain set of characteristics or assumptions, a value would take one of two distinct values.

Part (a); positive integers not exceeding 30.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the other twenty-four.

\frac{\left(\begin{array}{c}6 \\6\end{array}\right)\left(\begin{array}{c}24 \\0\end{array}\right)}{\left(\begin{array}{c}30 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}30 \\6\end{array}\right)}=1.68 \times 10^{-6}

Part (b); positive integers not exceeding 36.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the other thirty.

\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}30 \\0\end{array}\right)}{\left(\begin{array}{c}36 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}36 \\6\end{array}\right)}=5.13 \times 10^{-7}

Part (c); positive integers not exceeding 42.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the 36 other integers.

\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}36 \\0\end{array}\right)}{\left(\begin{array}{c}42 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}42 \\6\end{array}\right)}=1.91 \times 10^{-7}

Part (d); positive integers not exceeding 48.

To calculate the probability, use binomial coefficients. Choose six of the six accurate integers and none of the other 42.

\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}42 \\0\end{array}\right)}{\left(\begin{array}{c}48 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}48 \\6\end{array}\right)}=8.15 \times 10^{-8}

To know more about binomial probability, here

brainly.com/question/9325204

#SPJ4

The complete question is-

Find the probability of winning a lottery by selecting the correct six integers, where the order in which these integers are selected does not matter, from the positive integers not exceeding a) 30. b) 36. c) 42. d) 48.

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Which of the following equations has no solution?
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The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
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Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

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 = \frac{X - \mu}{\sigma}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

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Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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