What might you expect to happen if you first observed this constellation at 8:00pm tonight, then again at at 11:00pm?

7.3 as a mixed number simlified

joja [24]

Answer: 7 3/10
seven as a whole number, and .3 equals to 3/10 when converted

8 0

3 years ago

WILL MARK BRAINLIEST.

charle [14.2K]

Answer:

option d

Step-by-step explanation:

The cross section of the rectangular prism is parallel to the base.

So, the length and width is congruent to the base. So, same length and width

3 0

3 years ago

The Collatz conjecture is one of the most famous unsolved mathematical problems, because it's so simple, you can explain it to a

Vinil7 [7]

Hi, you've asked an incomplete question. However, I inferred you need a brief explanation about The Collatz conjecture.

<u>Explanation:</u>

Put simply, what the Collatz conjecture unsolved problem entails is that if any positive number is picked and it is:

An even number (eg 2, 4, 6,...), then if they are divided by 2, the new number gotten should undergo the same process (that is to be divided by 2), it is believed your calculation would finally end up at 1. For example, let's pick the number 6, (6÷2=4; repeating the process 4÷2=<u>1</u>)
An odd number, then if they are multiplied by 3 and 1 is added to the result, it is believed that your calculation would finally end up at 1.

6 0

3 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

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.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0

3 years ago

Alexa and her friends went out for pizza to celebrate the softball team’s victory. Their total bill for the pizza and drinks wa

Nitella [24]

Answer:

$39

Step-by-step explanation:

32.50:100 × 20 = $6.50

32.50+6.50 = $39

5 0

3 years ago