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elixir [45]
2 years ago
6

Solve the right triangle, ΔABC, for the missing sides and angle to the nearest tenth given angle B = 39° and side c = 13.

Mathematics
1 answer:
hram777 [196]2 years ago
5 0

Answer:

Part 1) b=8.2\ units

Part 2) a=10.1\ units

Part 3) A=51\° and C=90\°

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

Find the side b

we know that

In the right triangle ABC

The function sine of angle B is equal to divide the opposite side angle B (AC) by the hypotenuse (AB)

sin(B)=AC/AB

we have

AB=c=13\ units

AC=b

B=39\°

substitute

sin(39\°)=b/13

solve for b

b=(13)sin(39\°)

b=8.2\ units

step 2

Find the side a

we know that

In the right triangle ABC

The function cosine of angle B is equal to divide the adjacent side angle B (BC) by the hypotenuse (AB)

cos(B)=BC/AB

we have

AB=c=13\ units

BC=a

B=39\°

substitute

cos(39\°)=a/13

solve for a

a=(13)cos(39\°)

a=10.1\ units

step 3

Find the measure of angle A

we know that

In the right triangle ABC

C=90\° ----> is a right angle

B=39\°

∠A+∠B=90° ------> by complementary angles

substitute the given value

A+39\°=90\°

A=90\°-39\°

A=51\°

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Step-by-step explanation:

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Y2 ~ Bi (1,0.3)

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P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

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b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

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P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

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