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kompoz [17]
3 years ago
6

Jiao works as an usher at a theater. The theater has 1000 seats that are accessed through five entrances. Each guest should use

the entrance that's marked on their ticket. Jiao wants to test if the distribution of guests according to entrances matches the official distribution. He collects information about the number of guests that went through each entrance at a certain night. Here are the results: Entrance Expected # of people A 30% 398 B 30% 202 C 20% 205 D 10% 87 E 10% 108 Total 100% 1000 Jiao wants to perform a x' goodness-of-fit test to determine if these results suggest that the actual distribution of people doesn't match the expected distribution. # of people 398 202 205 87 108 1000 Jiao wants to perform a xa goodness-of-fit test to determine if these results suggest that the actua distribution of people doesn't match the expected distribution. What is the expected count of guests in entrance A in Jiao's sample? You may round your answer to the nearest hundredth.
Expected:______.
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Mathematics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

The answer is "300".

Step-by-step explanation:

Please find the complete solution in the attached file.

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D. 4.0<br> Geometry math question no Guessing and Please show work
worty [1.4K]

x=3.5

This is how I did it I took the bottom 2, 3 and 2.4 and I divided 3 by 2.4 (3/2.4) and got 1.25. I then multiplied 2.8 by 1.25 and found that x was 3.5

Answer is C. 3.5

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3 years ago
The series 10 + 17 + 24 + 31 + 38 + . . . is: arithmetic geometric both neither
Greeley [361]
The answer is geometric because its increasing by 7 each time
6 0
3 years ago
The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

slap that in your calculator and you get

t = 8.1 days

7 0
1 year ago
Soo I haven't done ratios in a while.. Anyone get this?
Sauron [17]
4/20 i think is the answerrr
4 0
2 years ago
Read 2 more answers
Given that f(x)=2x^3+x-3, evaluate the function f(-2). <br> A.) 11 <br> B.) -69 <br> C.) -21
prohojiy [21]
F(x)=2x³+x-3
f(-2)=2(-2)³+(-2)-3
f(-2)=2(-8)-2-3
f(-2)=-16-2-3
f(-2)=-18-3
f(-2)=-21. As a result,C is your final answer. Hope it help!
7 0
2 years ago
Read 2 more answers
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