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DiKsa [7]
3 years ago
7

In a survey of 100 outpatients who reported at a hospital one day,it was found that 70 complained of fever, 50 had stomach troub

le, and 30 were injured. All the 100 patientshad one or the other of these complaints, and 44 had exactly two of them. How many patients had all three complaints?
Mathematics
1 answer:
Verdich [7]3 years ago
6 0
6 patients had all three complaints.
We can't subtract 44 from 30.
we can subtract that from from 50 which leaves us with 6 patients that can have the third complaint as well.
We can also subtract 44 from 70 but the remaining number exceeds the number of patients with stomach ache.
so it leaves us with 6.
Revising:
So according to the survey
All 44 have exactly two complaints
30 patients were injured
50 had stomach ache
70 had fever
1. 30 is less than 44 so that leaves it outside
2. 50 is greater than 44 so we subtract 44 from 50. 6 patients are those who can have all three complaints.
3. 70 is also greater than 44 so we subtract 44 from 70. 26 patients are left behind.
The number of patients can't be higher than 6 because that is the number of patients that have stomach ache and can also have the other two complaints. Hence 6 patients have all three complaints.
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5 0
2 years ago
The sum of a number and its square is 182. Find the number.
Mars2501 [29]

Answer:

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Step-by-step explanation:

Let's represent that number as x

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So,

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x + x^2 - 182 = 182 - 182

x + x^2 - 182 = 0

rearrange the quadratic equation

x^2 + x -182 = 0

let's use the quadratic formula

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a = 1, b = 1, c = -182

\frac{-1+\sqrt{1^2-4*1*(-182)} }{2*1}       or     \frac{-1-\sqrt{1^2-4*1*(-182)} }{2*1}

\frac{-1+\sqrt{1+728} }{2}      or         \frac{-1-\sqrt{1+728} }{2}

\frac{-1+\sqrt{729} }{2}            or    \frac{-1-\sqrt{729} }{2}

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13 or    - 14

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8 0
3 years ago
A total of 8644 people went to the football game Of those people 5100sat on the home side and the rest sat on the visitors side
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Calculate the number of people who sat on the visitors' side:

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-5100
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Now divide 3544 people by 8 sections:

3544 people
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  8 sections
3 0
3 years ago
Read 2 more answers
Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa
4vir4ik [10]

Using the z-distribution, it is found that since the <u>test statistic is greater than the critical value</u>, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is <u>tested if the mean length of jail time is still of 2.5 years</u>, that is:

H_0: \mu = 2.5

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H_1: \mu > 2.5

We have the <u>standard deviation for the population</u>, thus, the z-distribution is used. The test statistic is given by:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

The parameters are:

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Hence, the value of the <u>test statistic</u> is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}

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A similar problem is given at brainly.com/question/24166849

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4 0
3 years ago
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