Pythagorean theorem !

(07.03 MC)

Umnica [9.8K]

The probability of pulling a blue marble and the coin landing tails up is 360/2500

<h3>How to determine the probability?</h3>

The tables of values are given as:

Color Times

Blue 18

Green 20

Yellow 12

Heads Tails

20 30

The probability of obtaining a blue marble is:

P(Blue) = 18/50

The probability of landing tails up is:

P(Tail) = 20/50

The required probability is:

P = P(Blue) * P(Tail)

This gives

P = 18/50 * 20/50

Evaluate the product

P = 360/2500

Hence, the probability of pulling a blue marble and the coin landing tails up is 360/2500

Read more about probability at:

brainly.com/question/25870256

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8 0

2 years ago

-2n( m^2 - m -3 ) =

kogti [31]

Answer:

$=-2nm^2+2nm+6n$

Step-by-step explanation:

$-2n\left(m^2-m-3\right)\\\mathrm{Distribute\:parentheses}\\=\left(-2n\right)m^2+\left(-2n\right)\left(-m\right)+\left(-2n\right)\left(-3\right)\\Apply\:minus-plus\:rules\\+\left(-a\right)=-a,\:\:\left(-a\right)\left(-b\right)=ab\\=-2nm^2+2nm+2\cdot \:3n\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6\\=-2nm^2+2nm+6n\\$

4 0

3 years ago

Read 2 more answers

Multiple-choice questions on Advanced Placement exams have five options: A, B, C, D, and E. A random sample of the correct choic

Semenov [28]

Answer:

Option B is not the most common correct choice.

Step-by-step explanation:

The multiple-choice questions on Advanced Placement exams have five options: A, B, C, D, and E.

The probability that any of these five option is the correct answer is:

$p=\frac{1}{5}=0.20$

A random sample of 400 multiple-choice questions on Advanced Placement exam are selected.

The results showed that 90 of the 400 questions having B as the answer.

To test the hypothesis that option B is more likely the correct answer for most question, the hypothesis can be defined as:

H₀: All the options are equally probable, i.e. p = 0.20.

Hₐ: Option B is more likely the correct option, i.e. p > 0.20.

Compute the sample proportion as follows:

$\hat p=\frac{90}{400}=0.225$

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.225-0.20}{\sqrt{\frac{0.20(1-0.20)}{400}}}= 1.25$

The test statistic is 1.25.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

$p-value=P(Z>1.25)\\=1-P(Z$

*Use a z-table.

The p-value is 0.1057.

The p-value of the test is quite large. Thus, the null hypothesis was failed to rejected.

Hence, it can be concluded that option B is not the most common correct choice.

7 0

2 years ago

A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this popul

Zina [86]

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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:

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

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.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$