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3 years ago

What is the first quartile (Q1) of the data set?

1 answer:

8 0

Sequence = 40, 45, 48, 51, 59, 62, 69, 74

Median = 51+59 / 2 = 55
First quartile = 40, 45, 48, 51
So, it would be: 45 + 48 / 2 = 46.5

In short, Your Answer would be Option C

Hope this helps!

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3 years ago

Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3/5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3root5 Find: KM

Sergio039 [100]

Answer:

The length of KM is $\sqrt{109}$ units.

Step-by-step explanation:

Given the statement: KLMN is a trapezoid, ∠N= ∠KML, FD=8, $\frac{LM}{KN}=\frac{3}{5}$ , F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, $ME=3\sqrt{5}$ .

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively.

The height of the trapezoid is $3\sqrt{5}$ .

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since $\frac{LM}{KN}=\frac{3}{5}$ , therefore LM is 3x and KN is 5x.

$\frac{3x+5x}{2}=8$

$\frac{8x}{2}=8$

$x=2$

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

$KN=KA+AE+EN$

$10=6+2(EN)$ (KA=EN, isosceles trapezoid) $EN=2$

$KE=KN-EN=10-2=8$

Therefore the length of KE is 8.

Use Pythagoras theorem is triangle EKM.

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Therefore the length of KM is $\sqrt{109}$ units.

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4 years ago