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3 years ago

60 twenty-fives minus 1 twenty-five

Mathematics

2 answers:

makkiz [27]3 years ago

5 0

60 twenty fives minus one twenty five is 59x25 is 1475.

9966 [12]3 years ago

5 0

Answer:

60 twenty fives minus 1 twenty five is:

59 twenty fives that is equal to 1425

Step-by-step explanation:

We need to know the difference between 60 twenty fives and 1 twenty five, so:

$60(25)-1(25)$

We have common factor, 25

$25(60-1)\\59(25)=1475$

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Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

Step-by-step explanation:

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

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Proportion of 0.6

This means that $p = 0.6$

Sample of 46

This means that $n = 46$

Mean and standard deviation:

$\mu = p = 0.6$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722$

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

$Z = \frac{0.5 - 0.6}{0.0722}$

$Z = -1.38$

$Z = -1.38$ has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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3 years ago

Explain how to add 678 + 303 using mental math

zloy xaker [14]

Answer:

You add it in your mind.

Step-by-step explanation:

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3 years ago

Multiply to find the product.<br><br> 64.7 x 28 = ______

Crazy boy [7]

Answer:

1811.6

Step-by-step explanation:

647x28

18116

move once to the left bc of the point in 64.7

1811.6

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2 years ago

Read 2 more answers

Based on the measures shown, could the figure be a parallelogram

nadya68 [22]

Answer:

What are the measures and is there a link that could be shown?

Step-by-step explanation:

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3 years ago

Data will be collected on the following variables. Which variable can be considered discrete? A. The height of a person B . The

kvv77 [185]

Answer: E: The number of books a person finished reading last month

Step-by-step explanation:

First, a discrete variable is a variable that only can take some given values in a set, the discrete variables are usually not dense, and a continuous variable is a variable that can take any value in a range (where the accepted values are dense).

So, for example, the set of the natural numbers is discrete, and the set of the real numbers can represent a continuous variable.

Here the only option that is really discrete will be the number of books that a person finished reading last month because here only positive whole numbers are accepted (you can not finish a 0.454 of a book)

The other options are continuous because all are classical measures.

For example, the weight of a person can jump between:

75.6kg and 75.7kg.

So you could think that this is discrete because the values between 75.6kg and 75.7g are not shown with our measuring device, but those will be added in the error of the measure because the weights between 75.6kg and 75.7kg are actually possible, so they must be accepted.

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3 years ago