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inna [77]
3 years ago
8

WORTH 15 POINTS PLEASE ANSWER ASAP! IF IT LETS ME MARK YOU BRAINLIST I WILL!

Mathematics
2 answers:
masya89 [10]3 years ago
8 0

Answer:

the answer is 3/8

Step-by-step explanation:

multiply 9/10 by 5/12 then you get the answer 3/8

Debora [2.8K]3 years ago
6 0
Answer is 3/8 good job have a nice day
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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}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

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

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

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}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

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

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}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

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

7 0
3 years ago
Express the set of numbers shown on the number line
laiz [17]

Given:

The graph of a number line.

To find:

The set of values for the given number line.

Solution:

As we move from right to left on a number line the values decreased however as we move from left to right on a number line the values increased.

From the given number line it is clear that there is an open circle at 7  and a close circle at 10. It means 7 is not included in the solution set but 10 is included in the solution set.

From the point 7, the arrow approaching to the left it means all the values less than 7 are included in the solution set, i.e., x < 7.

From the point 10, the arrow approaching to the right it means all the values greater than 10 or equal to 10 are included in the solution set, i.e., x ≥ 10.

Solution set : x < 7 or  x ≥ 10.

Therefore, the correct option is A.

3 0
3 years ago
How do you solve 14g &gt; 56​
Alona [7]

You divide both sides by 14:

14g>56 \iff \dfrac{14g}{14}>\dfrac{56}{14} \iff g>4

4 0
4 years ago
Mutltply 2/3 by 42 and then mutltply that product by 10
olga2289 [7]

Answer:

280

Step-by-step explanation:

2/3 * 42 = 28

28 * 10 = 280

6 0
3 years ago
Read 2 more answers
Solve for v 209=40-v
andrew-mc [135]

Answer:

-169

Step-by-step explanation:

We want to isolate v.

Method 1:  subtract 40 from both sides, obtaining  169 = -v.  Then change the sign of both sides, arriving at v = -169.

Method 2:  Add v to both sides.  Then v + 209 = 40.

Now subtract 209 from both sides, obtaining -169.

6 0
3 years ago
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