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damaskus [11]
3 years ago
15

I'm sorry I suck at this triangle stuff. How many unique triangles can be made where one angle measures 60° and another angle is

an obtuse angle?
A.) 1
B.) 2
C.) More than 2
D.) None
Mathematics
2 answers:
Paha777 [63]3 years ago
7 0

Answer:

I think it's more than 2 because some examples are 60,91,29 and 60, 92, 28 and 60, 93, 27.

igor_vitrenko [27]3 years ago
4 0

Answer:

C

Step-by-step explanation:

60+x+y = 180

x > 90

There are more than 2 possiblities

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Evan has a storage unit that he keeps all his garden tools in. It measures 4 feet by 312 feet by 2 feet.A prism has a length of
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The maximum amount of supplies that the storage unit can hold is 28 ft³

<h3>How to calculate the volume of a rectangular prism?</h3>

For us to calculate the volume or amount of space in Evan's prism, we will first of all calculate the volume of rectangular prisms. Formula is;

V = Length × width × height.

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6 0
2 years ago
A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.
AVprozaik [17]

Answer:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean  is also normal and given by:

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\mu_{\bar X} = 61

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So then is appropiate use the normal distribution to find the probabilities for \bar X

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 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".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Solution to the problem

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

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean \bar X is also normal and given by:

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

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

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