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Anna [14]
3 years ago
12

Simon draws a rectangle with two sides that are 2 inches long and two sides that are 3 inches long. Chaz draws a figure with the

same measurements, but it is not a rectangle. What figure could it be? i need the answer this is due Tomorrow. Please give me the answer. SOMEONE PLEASE HELP.
Mathematics
1 answer:
Lapatulllka [165]3 years ago
8 0
Chad could've made a mistake and drew an acute triangle, thats a triangle that has 2 sides thats 3 inches and 1 side dats 2 inches
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The final amount is 40% of the original (100%-60%=40%).
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How many outcome sequences are possible when a die is rolled 4 times where we say for instance that the outcome is 3 4 3 1 if th
STatiana [176]

Answer:

the possible outcome sequences when a die is rolled 4 times is 1296

Step-by-step explanation:

 Given the data in the question;

a die is rolled 4 times

and outcomes are { 3, 4, 3, 1 }

we know that; possible number of outcomes on a die is n = 6{ 1,2,3,4,5,6 }

Now when we roll a die lets say, r times

then the total number of possible outcomes will be;

N = n^r

given that; r = 4

Hence if we roll a die 4 times;

Total number of possible outcome N = 6⁴

N = 1296

Therefore, the possible outcome sequences when a die is rolled 4 times is 1296

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3 years ago
I don't understand the answer to the problem
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you are going in vacation you want to visit 4 different places in 2 weeks and you don't care what order in which you visit them.
ankoles [38]

Answer:

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Step-by-step explanation:

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7 0
2 years ago
The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma
nata0808 [166]

Answer:

a) \sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b) Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

c) P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d) P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e) P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

X \sim N(110000,40000)  

Where \mu=110000 and \sigma=40000

If we select a sample size of n =35 the standard error is given by:

\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

P(X > 112000)

And we can use the z score given by:

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

And replacing we got:

P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

P(X > 100000)

And we can use the z score given by:

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

And replacing we got:

P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

P(100000

And we can use the z score given by:

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

And replacing we got:

P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

8 0
4 years ago
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