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garri49 [273]
3 years ago
15

Below, the two-way table is given for a

Mathematics
1 answer:
statuscvo [17]3 years ago
6 0

Answer:

40%

Step-by-step explanation:

Probability that student is a male Given that it's a senior ::

Recall :

P(A | B) = (AnB) / B

Hence, we have ;

P(male | Senior) = number of (Male n Senior) / total Senoirs

From the table :

Total seniors = 2 + 3 = 5

Male n Seniors = 2

P(male | Senior) = 2 / 5 = 0.4 = 40%

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There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t
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Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s  text which one of the triangle's  side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM.  AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM  (1)

1. Let AC=21   So AM=21/2=10.5 cm

So AB=10.5 cm as well.  So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill.  AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So  BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB.  So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC    21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

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Step-by-step explanation:

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