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arsen [322]
4 years ago
6

Kwame's team will make two triangular pyramids to decorate the entrance to the exhibit. They will be wrapped in the same metalli

c foil. Each base is an equilateral triangle. If the base has an area of about 3.9 square feet, how much will the team save altogether by covering only the lateral area of the two pyramids? The foil costs $0.24 per square foot. Kwame's team will save $
, altogether by covering only the lateral area of the two pyramids.
Mathematics
1 answer:
Veseljchak [2.6K]4 years ago
8 0

Answer:

Kwame's team will save = 7.8 \times $0.24 = $1.87

Step-by-step explanation:

i.) Let the side of the equilateral triangle base be a

ii.) the area of the base = 3.9 square feet

iii.) the area of equilateral triangle is = \frac{\sqrt{3} }{4} a^{2} = 3.9

iv.) Base area  = 3.9 square feet

v.) The area that is not covered is the base.

vi.) The total area that is not covered  = 3.9 \times 2  since there are two pyramids

    therefore the total area not covered = 7.8 square feet

vii.) therefore Kwame's team will save = 7.8 \times $0.24 = $1.87

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Answer:

Grade A: Z \geq 1

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Step-by-step explanation:

Problems of normally distributed samples can be solved using the Z score table.

The Z score of a measure represents how many standard deviations it is above or below the mean of all the measures.

Each Z score has a pvalue. This represents the percentile of the measure.

In this problem, we have that:

The upper 16% of the class get A grades. The upper 16% has a pvalue of at least 100% = 16% = 84% = 0.84. This is Z \geq 1.

The middle 34% of the class get B grades. The middle 34% has a pvalue of at least 84%-35% = 50% = 0.5 and at most 0.84. This is 0\leqZ < 1.

Those between a pvalue of 0.5-0.34 = 0.16 and 0.5 get get grade C. Z = -1 has a pvalue of 0.16. So a grade C is in the interval -1 \leq Z < 0.

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3 years ago
2^2 times 2^3<br> a.2^4<br> b.2^5<br> c.2^6<br> d.4^6
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Answer:

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4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

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2x² - 8x - 3, here a = 2, b = -8, c = -3.                                                    x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4.     88 can also be written as 4 × 22 and √4 = 2. So                                                                             x = 8 ± 2√22 / 2×2 = 4± √22/2.

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2x² - 9x + 6, here a = 2, b = -9, c = 6.                                                    x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4.                                                                             x = 9 ± √33 / 4 .

5 0
3 years ago
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