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Tomtit [17]
4 years ago
8

The P.E. teacher at Sunshine Middle School wants to know which sports are popular among students in her school. Since it would t

ake so long to ask all of the students, she will survey a group of students. The school has 288 6th graders, 256 7th graders, and 280 8th graders. The P.E. teacher surveys 36 6th graders and 32 7th graders. How many 8th graders should she survey in order for the group to be a representative sample?
Mathematics
1 answer:
sineoko [7]4 years ago
5 0
28 BECAUSE IT CONTINUES THE PATTERN OF SUBTRACTING 4 FROMTHE AMOUNT PULLED PER GRADE
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There was 2/3 of a pan of lasagna in the refrigerator. Bill and his friends ate 1/2 of what was left. How much of the pan did th
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Don't answer it if you don't know
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3 years ago
Write the equation of the line passing through points (2,-2) and (1,0) in the slope intercept form​
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7 0
3 years ago
A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude ​
I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

  • ( a ) = 35 cm

  • ( b ) = 54 cm

  • ( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

  • <u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

\sf{   \pmb { \longrightarrow \:  \sqrt{s(s - a)(s - b)(s - c)} }}

where ( s ) :

\sf \longrightarrow s = \dfrac{a + b + c}{2}

Therefore, for the given triangle first we will calculate ( s )

\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

Now, Area of triangle will be:

\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

3 0
2 years ago
Read 2 more answers
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