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maksim [4K]
3 years ago
7

A stadium can seat 46,000 people for a baseball game. One day, 49,442 people attended a game. How many people had to stand becau

se they did not have a seat? A. 9,442 B. 3,442 C. 4,442 D. No one stood; there were enough seats
Mathematics
2 answers:
suter [353]3 years ago
5 0

Answer:

The answer is B.

Step-by-step explanation:

So, there are 46,000 seats for a baseball game and 49,442 attended. You need to subtract 46,000 from 49,442 to get the answer that is 3,442.

ollegr [7]3 years ago
5 0
The answer is B: 3,442
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A recent survey of 128 high school students indicated the following information about release times from school.
Delicious77 [7]

Answer:

A) Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

B) The group that wants to get out of school earlier can win after all the votes are counted if their true population proportion takes on a value that is higher than the closest true population proportion (for the group that wants to get out of school later)

That is, in the (31%, 39%) and (35%, 43%) obtained in (a), a range of (35.1%, 39%) and (35%, 38.9%) show how possible that the group that wants to get out of school earlier can win after all the votes are counted.

Step-by-step explanation:

The Interval estimate for the proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Interval estimate = (Sample proportion) ± (Margin of error)

Margin of Error is the width of the confidence interval about the proportion.

a) Find the interval estimates

i) for those that want to get out earlier and

ii) those that want to get out later.

i) Sample proportion of those that want to get out earlier = 35%

Margin of Error = 4%

Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

ii) Sample proportion of those that want to get out later = 39%

Margin of Error = 4%

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

b) Explain how it would be possible for the group that wants to get out of school earlier to win after all the votes are counted.

Since the interval estimates represent the range of values that the true population proportion can take on for each group that prefer a particular option, the group that wants to get out of school earlier van have their proportion take on values between 31% and 39%. If their true population takes on a value that is highest (which is very possible from the interval estimate), and the group with the highest proportion in the sample, (the group that wants to get out of school later, whose true population proportion can take between 35% and 43%) has a true population proportion that is less than that of the group that wants to get out of school earlier, then, the group that wants to get out of school earlier can win after all the votes are counted.

Hope this Helps!!!

3 0
4 years ago
The table shows the total profits earned from magazine sales.
coldgirl [10]

Answer:

C. y= 14x+20

Step-by-step explanation:

By looking at the table, you can put in the x-values into all of the equations until you find the corresponding y-values that matches with the values in the table. For example: y= 14(2)+ 20, y=48. y= 14(20) +20, y=300. Hope this helped!

8 0
3 years ago
Read 2 more answers
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

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3 years ago
Joseph is going to paint the panels of a fence. He is going to use the same repeating pattern for the entire fence. The rule is
Marianna [84]

19 times Answer:

Step-by-step explanation:

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3 years ago
in a sale there is 25% off and a bed costs 33 pounds in the sale so how much was it before the sale ?
IceJOKER [234]
44 pounds 
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