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Tomtit [17]
2 years ago
6

A special 8-sided die is marked with the numbers 1 to 8. It is rolled 20 times with these outcomes:

Mathematics
2 answers:
Advocard [28]2 years ago
8 0

Answer:

One: C

Two: 60% and 10%

Step-by-step explanation:


horsena [70]2 years ago
4 0

Answer:

One: B

Two: 60% and 10%

Step-by-step explanation:

Problem One

There are only two numbers in the sample of 40 that are under 26. Both are 25. If you find more, make the adjustment. There are 2 more that are exactly 26 but they are not counted because the directions say "less than 26."

So set up your proportion

x/2000 = 2/40                     Multiply both sides by 2000

x = 2/40 * 2000

x = 4000/40

x = 100

A

I don't know where 5 comes from. But it is not correct.

B

B should be the correct answer.

C

Exactly 100 pieces should be defective. That is the theoretical result. C is incorrect.

D

D is not correct. The sample size would not be 40. It would have to be 2000 for D to be correct. So D is wrong.

E

We have enough data to get an answer. E is incorrect.

Problem 2

The think you must NOT do is count 1 as being prime. The prime numbers are 2 3 5 7 between 1 and 8. They break down as follows.

  • Prime           Number of them
  • 2                         3
  • 3                         4
  • 5                         2
  • 7                         3

The total number of primes = 12

There are 20 numbers in the sample

The experimental probability of tossing a prime is 12/20 * 100% = 60%

The non primes are 2 3 5 7 which is 4 out of 8

4/8 * 100 = 50%

The experimental value is 10% more than the theoretical value.

Discussion

Note: the problem may be one. This all depends on what you have been told about 1. I am using the exact wording of prime here. 1 is not a prime. It is also not a composite. So it has to be counted as part of the non primes.


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The equation would be x+5*-2=6.

The result of that equation would be x=16
7 0
2 years ago
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Solve the system of linear equations using multiplication.
Anna71 [15]

Answer:

(8,-1)

Step-by-step explanation:

The given system is:

3x+3y=21

6x+12y=36

Since I prefer to use smaller numbers I'm going to divide both sides of the first equation by 3 and both sides of the equation equation by 6.

This gives me the system:

x+y=7

x+2y=6

We could solve the first equation for x and replace the second x with that.

Let's do that.

x+y=7

Subtract y on both sides:

x=7-y

So we are replacing the second x in the second equation with (7-y) which gives us:

(7-y)+2y=6

7-y+2y=6

7+y=6

y=6-7

y=-1

Now recall the first equation we arranged so that x was the subject. I'm referring to x=7-y.

We can now find x given that y=-1 using the equation x=7-y.

Let's do that.

x=7-y with y=-1:

x=7-(-1)

x=7+1

x=8

So the solution is (8,-1).

We can check this point by plugging it into both equations.

If both equations render true for that point, then we have verify the solution.

Let's try it.

3x+3y=21 with (x,y)=(8,-1):

3(8)+3(-1)=21

24+(-3)=21

21=21 is a true equation so the "solution" looks promising still.

6x+12y=36 with (x,y)=(8,-1):

6(8)+12(-1)=36

48+(-12)=36

36=36 is also true so the solution has been verified since both equations render true for that point.

5 0
2 years ago
Read 2 more answers
(a) Suppose one house from the city will be selected at random. Use the histogram to estimate the probability that the selected
vodka [1.7K]

Answer:

a.    0.71

b.    0.9863

Step-by-step explanation:

a. From the histogram, the relative frequency of houses with a value less than 500,000 is 0.34 and 0.37

-#The probability can therefore be calculated as:

P(500000)=P(x=0)+P(x=500)\\\\\\\\=0.34+0.37\\\\\\\\=0.71

Hence, the probability of the house value being less than 500,000 is o.71

b.

-From the info provided, we calculate the mean=403 and the standard deviation is 278 The probability that the mean value of a sample of n=40 is less than 500000 can be calculated as below:

P(\bar X

Hence, the probability that the mean value of 40 randomly selected houses is less than 500,000 is 0.9863

8 0
2 years ago
Company F sells fabrics known as fat quarters, which are rectangles of fabric created by cutting a yard of fabric into four piec
jeka94

Answer:

a) Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean= 0.45, S.D= 0.6718

c) mean= 1.285, S.D= 8.74

Step-by-step explanation:

a) The following table shows the probability distribution of X:

X 0 1 2 3 4 or more

P(X) 0.58 0.23 0.11 0.05 0.03

Defect >2 = cannot be sold

Y = the number of defects on a fat quarter that can be sold by Company F.

Y = defect that can be sold

Y = Defect less or equal to 2 = 0,1,2

Probability distribution of the random variable Y:

Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean of Y (μ)

μ = Σ x*P(Y)

= (0*0.58) +(1*0.23)+(2*0.11)

= 0+0.23+0.22 = 0.45

Standard deviation of Y = σ

σ = Σ√(x-mean)^2*P(Y)

= Σ√[(x- μ )^2*P(Y)]

= √[(0-0.45)^2*0.58+ (1-0.45)^2*0.23 + (2-0.45)^2*0.11]

= √[0.11745 + 0.069575 +0.264275

= √(0.4513

σ = 0.6718

Company G:

σ for defect that be sold = 0.66

μ for defect that be sold = 0.40

Difference between μ of F and μ of G

= 0.45-0.40 = 0.05

Difference between σ of F and σ of G

= 0.67-0.66 = 0.01

Selling price of fat quarter without defect = $5

Discount per defect = $1.5

Selling price per defect = 5-1.5 = $3.5

Discount per 2 defect = $1.5*2 = $3

Selling price per defect = 5-3 = $2

Since defect to be sold cannot be greater than 2, let Y = 5,3,2

Probability distribution of the selling price Y:

Y 5 3 2

P(Y) 0.58 0.23 0.11

μ = (5*0.58) +(3.5*0.23)+(2*0.11)

μ = 2.9+0.805+0.22 =1.285

σ = Σ√[(x- μ )^2*P(Y)]

σ = √[(5-1.285)^2*0.58+ (3-1.285)^2*0.23 + (2-1.285)^2*0.11]

σ = 8.00+0.68+0.06 = 8.74

7 0
2 years ago
A map of Colorado says that the scale is 1 inch to 20 miles or 1 to 1,267,200. Are these two ways of reporting the scale the sam
goldenfox [79]

Answer:

1,267,200

Step-by-step explanation:

Given that the map reported that the scale is 1 inch to 20 miles.

Recall that there are 63,360 inches in 1 mile, thus, in 20 miles, we have 20 x 63,360 = 1,267,200

Therefore, the scale of 1 inch to 20 miles can also be represented as 1 to 1,267,200

4 0
1 year ago
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