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AfilCa [17]
3 years ago
13

To prepare for virtual learning the school is purchasing extra keyboards, mice, and PC cameras. They have a budget of $1500 to s

pend on $30 keyboards, $20 mice, and $50 cameras. Additionally, the number of mice should be equal to that of keyboards and twice the number of cameras. How many of each item should he buy?
explain please​
Mathematics
1 answer:
leonid [27]3 years ago
7 0

Answer:

20 keyboards and mice, 10 cameras.

Step-by-step explanation: They need the same number of keyboard and mice, and they want to get half the amount of cameras they do keyboard and mice. They have a budget of $1500 so 20 keyboards and mice as well as 10 cameras is the most optimal purchase.

30*20=600

20*20=400

50*10=500

600+400+500=1500

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At a certain car dealership, the probability that a customer purchases an SUV is . Given that a customer purchases an SUV, the p
SpyIntel [72]

Answer:

The probability that a customer purchases a black SUV is 0.05.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>At a certain car dealership, the probability that a customer purchases an SUV is </em><em>0.20</em><em>. Given that a customer purchases an SUV, the probability that it is black is </em><em>0.25</em><em>.</em>

The probability that a customer purchases a black SUV can be calculated as the multiplication of this 2 factors:

  • The probability of a customer purchasing a SUV: P(SUV).
  • The probability that it is black, given that he or she purchases a SUV (conditional probabilty): P(B|SUV)

We know then:

P(SUV)=0.25\\\\P(B | SUV)=0.20

We can now calculate the probability as:

P(B\,\&\,SUV)=P(B|SUV)\cdot P(SUV)=0.25\cdot0.20=0.05

3 0
3 years ago
CAN ANYONE HELP ME PLEASE 20 POINT!!!
cricket20 [7]

Answer:

the answer should be B -3

Step-by-step explanation:

1+3-5-2= -3

6 0
3 years ago
Read 2 more answers
It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.
zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

P(X=x)=nCx*p^{x}*(1-p)^{(n-x)} with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability  

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)} with x = 0, 1

So,

P(A)=P(X=0)+P(X=1)

P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}

P(A)=0.99^{10}+10*0.01*0.99^{9}

P(A)=0.9957

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127

Finally, we have that the probability of returning exactly one of the three packages is 1.27%

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3 years ago
What is the value of x in the picture ?
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Suppose we have the following information:
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A- The probability that it will rain today or tomorrow
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