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3 years ago

14

How much material is needed to construct a triangular tent that is 10 feet wide, 12 feet tall, and 18 feet long with side measur

ements of 13 feet?
768 square feet

1024 square feet

1120 square feet

1392 square feet

1 answer:

Ket [755]3 years ago

3 0

The answer is " 768 square feet. ".

Total surface area of a triangular prism = width x height +(sum of sides + width) x length and here;
width = 10
height = 12
sides = 13
length = 18
= (10 x 12) + (13 + 13 + 10) x 18
= 120 + (36 x 18)
= 120 + 648
= 768 square feet.

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Step-by-step explanation:

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Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

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$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

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4. West point from the previous: By the 1st theorem,

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5. West point from the previous: By the 2nd theorem,

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$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

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$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

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And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

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[1] 28.32

And the interval would be (25.72; 28.32)

Step-by-step explanation:

Notation and definitions

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$s=5.83$ represent the sample standard deviation

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A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 88% of confidence, our significance level would be given by $\alpha=1-0.88=0.12$ and $\alpha/2 =0.06$ . The degrees of freedom are given by:

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[1] -1.582366

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[1] 1.582366

The critical value $tc=\pm 1.582366$

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The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=1.582366 \frac{5.83}{\sqrt{50}}=14.613$

With R we can do this:

m=1.582366*(5.83/sqrt(50))

m

[1] 1.304639

Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$27.02 - 1.582366 \frac{5.83}{\sqrt{50}}=25.715$

$27.02 + 1.582366 \frac{5.83}{\sqrt{50}}=28.325$

Using R the code is:

lower=27.02-m;lower

[1] 25.71536

upper=27.02+m;upper

[1] 28.32464

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

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[1] 28.32

And the interval would be (25.72; 28.32)

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