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S_A_V [24]
3 years ago
13

During the settlement of the west many towns were established by speculators who

Mathematics
1 answer:
USPshnik [31]3 years ago
8 0

Answer:

could you finish the problem?

Step-by-step explanation:

Also this is in the wrong section

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If a cat falls out of a tree, it has a probability of 0.92 that it will land
Alex_Xolod [135]

The probability that the cat lands on its feet both times it falls out of a tree is 0.85

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

The probability that the cat lands on its feet is given as:

P(Feet) = 0.92

The probability that it lands on its feet twice is calculated using:

P = P(Feet) * P(Feet)

So, we have:

P = 0.92 * 0.92

Evaluate

P = 0.85

Hence, the probability is 0.85

Read more about probability at

brainly.com/question/251701

#SPJ1

3 0
2 years ago
Please solve with explanation
valkas [14]

Answer:

a

Step-by-step explanation:

5 0
3 years ago
The difference between two integers is 13 and their sum is 87. What are the two integers
e-lub [12.9K]
The answer would be 50 and 37.

50-37 = 13

50+37 = 87
4 0
3 years ago
Read 2 more answers
An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
Ms Hart had 40 dollars to spend on costume boxes for the musical. If she bought 12 boxes and received 4 dollars in change, how m
artcher [175]

Answer:each box cost =$3

Step-by-step explanation:

Amount of money Ms Hart has to spent on costume boxes=$40

Amount gotten as change after purchase=$4

Amount actually used up buying costume boxes =$40-$4=36

Since she bought 12 boxes , each box cost =$36/ 12=$3

5 0
3 years ago
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