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LuckyWell [14K]
2 years ago
9

What is the length of the hypotenuse of the triangle?

Mathematics
2 answers:
masha68 [24]2 years ago
8 0
10 star-root 2 end units
Kaylis [27]2 years ago
4 0

Answer: 5 StartRoot 2 EndRoot Units

Step-by-step explanation:

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A triangle has an angle that measures 90 degrees what type of triangle could it be​
vekshin1

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

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Given m||n, find the value of x.<br><br> (2x+9)<br> (7X+24)
motikmotik

<em>Given, 9−7x=5−3x</em>

<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.</em>

<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x</em>

<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x⇒4=4x</em>

<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x⇒4=4x⇒x= </em><em>4</em><em>/</em><em>4</em><em> </em><em> =1</em>

<em> =1Hence, required solution is x=1.</em>

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2 years ago
A line has a slope of 5 and passes through the point
podryga [215]

Answer:

Step-by-step explanation:

The equation is mx+b=y.

M=slope (in this case 5)

x=x coordinates

b=y intercept

y=y coordinates

For the x and y coordinates, I'll use (2,0).

5(2)+_=0

10+_=0.

In this case, we only have one awnser, -10. Since 10+(-10) are the only ways to equal 0.

Hope this helps,

5 0
3 years ago
Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace
joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

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

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

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