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Anarel [89]
2 years ago
6

Find the missing term. () + (7 − 3i) + (5 + 9i) + 13i = 10 − 5i.

Mathematics
2 answers:
kotegsom [21]2 years ago
5 0
X + (7 - 3i) + (5 + 9i) + 13i = 10 - 5i

Subtract 13i from both sides

x + (7 -3i) + (5 + 9i) = 10 - 18i

Subtract (5 + 9i). MAKE SURE YOU SUBTRACT 9i TOO. In other words, distribute the negative and subtract 5 and 9i at the same time.

x + (7 - 3i) = 5 - 27i

Do the same with (7 - 3i). You'll be adding 3i since -(-3i) = 3i.

x = -2 - 24i
Archy [21]2 years ago
3 0

Answer:

the missing term is (-2-24i)

Step-by-step explanation:

Let

(x+yi)------> the missing term

we have

(x+yi)+(7-3i)+(5+9i)+13i=10-5i

Combine like terms in the left side

(x+7+5)+(yi-3i+9i+13i)=10-5i

(x+12)+(yi+19i)=10-5i

Solve the system of equations

x+12=10 ------> equation A

x=10-12=-2

yi+19i=-5i ------> equation B

yi=-5i-19i=-24i

the missing term is (-2-24i)

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Answer:

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2 years ago
A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L
Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

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nCx=\frac{n!}{x!(n-x)!}

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In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

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Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

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

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

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

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frozen [14]
Using sin rule:
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SIN A=sin 46=0.72
THEN:sin B =(0.72*26)/33=0.57
then B= 34° 
AS the sum of triangle angles=180 
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then c =(a*sin C)/sin A
sin C=sin 100=0.98
=(33*0.98)*0.72=45
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