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lana [24]
3 years ago
6

If a triangle has side lengths 7, 10, and 12, is it a right triangle??

Mathematics
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

No. Remember, a right angle must have a 90 degree angle. We can find the lengths with the Pythagorean Theorem.

Step-by-step explanation:

Given the length 7, 10, and 12, we can assume that 12 is the hypotenuse (it is the longest length).

- we can use 7 and 10 interchangeably.

Fill in the equation, a^2 + b^2 = c^2

                 where c = 12, and a or b = 7 or 10.

To indicate if the given lengths would form a right angle, we can only input 7 or 10, not both.

Therefore, 7^2 + b^2 = 12^2 or 10^2 + b^2 = 12^2

7^2 + b^2 = 12^2 ==> 49 + b^2 = 144 ==> <u>b= </u>\sqrt{95<u> ==> </u><u>9.746</u>

b= 9.7, not 10.

10^2 + b^2 = 12^2 ==> 100 + b^2 = 144 ==> <u>b = </u>\sqrt{44<u> ==> </u><u>6.633 </u>

b= 6.6, not 7.

Therefore, the lengths 7, 10, and 12, does NOT make a right triangle.

Hope this helps!

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3 years ago
A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

            \sigma = population standard deviation = 0.001 g

            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

                                             = [ 3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } } , 3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } } ]

                                             = [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0
3 years ago
(x+5)(x+3) Polynomial in standard form<br>​
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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