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Yuri [45]
2 years ago
11

Sarah has gone to work for the past 60 days. On 39 of those days she arrived at work before 8 AM. On the rest of those days she

arrived after 8:30 AM. What is the experimental probability that she will arrive at 8 AM on the next day she goes to work.
Mathematics
1 answer:
puteri [66]2 years ago
4 0

Answer:

13/20

Step-by-step explanation:

Probability is the ratio of the number of possible outcome to the number of total outcome.

Given that for the past 60 days, on 39 of those days she arrived at work before 8 AM while for the rest of those days she arrived after 8:30 AM, the probability that she will arrive at 8 AM on the next day she goes to work

= 39/60

= 13/20

In decimal, this will be expressed as 0.65 and 65% in percentage.

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What is the difference in length between the longest meteor and the shortest meteor?
hichkok12 [17]
1 5/8
1 7/8 - 2/8= 8+7/8 - 2/8= 15/8 - 2/8= 13/8=
1 5/8 is the answer
6 0
3 years ago
Consider the function f(x)=-2x+5 what is f(5) enter your answer in the box
Oliga [24]

The value of f(5) is f(5)=-5

Step-by-step explanation:

The function is f(x)=-2x+5

To find f(5) substitute x=5 in f(x)=-2x+5

f(5)=-2(5)+5

Multiplying the terms, we get,

f(5)=-10+5

Adding the terms, we get,

f(5)=-5

Thus, the value of f(5) is f(5)=-5

8 0
3 years ago
Which quadrant does the coordinate (-8, 12) lie?
marysya [2.9K]
In quadrant 2 is where it lies.
3 0
2 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
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KATRIN_1 [288]
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The others result in a lower number than what they are supposed to be. The 3 correct ones end with the exact number of the equations solution number.
7 0
2 years ago
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