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yKpoI14uk [10]
2 years ago
12

Two students compared the scores on their math tests. Their results on 9 tests are shown in the box plots below.

Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
4 0

The statements that are true for the box plots are:

C. The median of Nadia’s data = median of Ben’s data.

D. Nadia had the highest score on a test

<h3>How to Find the Median of a Box Plot?</h3>

The median of a data set is the center value, which is indicated on a box plot by a vertical line that divides the box.

Given the two box plots, we can conclude that:

Median for the data of Nadia is 92, which is the same for Ben.

The extreme whisker to the right for Nadia is at 100, which is the highest value for Nadia while that of Ben is 99.

The true statements are:

C. The median of Nadia’s data = median of Ben’s data.

D. Nadia had the highest score on a test

Learn more about the median on a box plot on:

brainly.com/question/14277132

#SPJ1

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Find the intervals on which the function is increasing or decreasing.
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To solve this, you have to know that the first derivative of a function is its slope. When an interval is increasing, it has a positive slope. Thus, we are trying to solve for when the first derivative of a function is positive/negative.

f(x)=2x^3+6x^2-18x+2
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1. Mr. Hurst's house increased in value to $185,400 one year after he bought
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NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
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Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

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