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ahrayia [7]
3 years ago
10

A triangle has an angle that measures 90 degrees what type of triangle could it be​

Mathematics
2 answers:
vekshin13 years ago
6 0

A triangle that has an angle that measures 90 degrees would be a Right Triangle.

Dimas [21]3 years ago
5 0

That would be a right triangle!

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How to simplify this
algol [13]
To simplify this, you would have to turn b^-2 into a positive exponent.
To do this, we have to flip b^-2, which would get rid of the negate from the exponent: -2

3a^4 b^-2 c^3 / b^-2

Then we get the answer:

3a^4 c^3
------------
   b^-2

I have a picture to clarify! 
I hope this helped, let me know if you don't understand! ^.^

6 0
2 years ago
$27.99 shoes ; 7 1/2 tax find the total cost or sale price to the nearest cent
Vinvika [58]
27.99x.075= 2.09925 or 2.10 Rounded
27.99+2.10= $30.09

5 0
3 years ago
Read 2 more answers
1.Amulya is x years now. Write expressions for her age 5 years ago?
ikadub [295]

Answer:

Step-by-step explanation:

1)    a = x - 5

a)   z = y + 5

b)  z = y - 3

c)   z = 6y

d)  z = 6y - 2

e) z = 3y + 5

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2 years ago
Help!! I hate geometryyyy
miskamm [114]

Answer:

AFB, BDC, and FBD

Step-by-step explanation:


3 0
3 years ago
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is per night (USA Today, Apr
kow [346]

Answer:

P(X

Step-by-step explanation:

Assuming a mean of $204 per night and a deviation of $55.

a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?

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 cost per night at the hotel, and for this case we know the distribution for X is given by:

X \sim N(204,55)  

Where \mu=204 and \sigma=55

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

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

We are interested on this probability

P(X>225)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>225)=P(\frac{X-\mu}{\sigma}>\frac{225-\mu}{\sigma})

=P(Z>\frac{225-204}{55})=P(Z>0.382)

And we can find this probability on this way:

P(Z>0.382)=1-P(Z

8 0
2 years ago
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