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NeX [460]
3 years ago
12

in the year 2006 there were 163,476 wildfires in a county. This was 12520 less than four times the number of wildfires in the ye

ar 2003. How many wildfires were there in 2003?
Mathematics
1 answer:
sesenic [268]3 years ago
4 0

x= wildfires in 2003

163,476= 4x - 12,520

(163,476 + 12,520) / 4

x = 43,999 your answer is 43,999

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The parent volunteers made 120 snow cones in 5 hours. The student council made 100 snow cones in 4 hours. Who made more snow con
marysya [2.9K]
The parent volunteers:
120 snow cones ... 5 hours
x snow cones = ? ... 1 hour

120 * 1 = 5 * x
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x = 120 / 5 = 24 snow cones

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A researcher is concerned about the impact of students working while they are enrolled in classes, and she likes to know if stud
8_murik_8 [283]

Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(c) Margin of error = 0.800

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

\bar x = 7.10 -- sample mean

\sigma=5 --- sample standard deviation

n = 150 --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

CI = 90\% --- the confidence interval

Calculate the \alpha level

\alpha = 1 - CI

\alpha = 1 - 90\%

\alpha = 1 - 0.90

\alpha = 0.10

Divide by 2

\frac{\alpha}{2} = 0.10/2

\frac{\alpha}{2} = 0.05

Subtract from 1

1 - \frac{\alpha}{2} = 1 - 0.05

1 - \frac{\alpha}{2} = 0.95

From the z table. the critical value for 1 - \frac{\alpha}{2} = 0.95 is:

z = 1.960

Solving (c): Margin of error

This is calculated as:

E = z * \frac{\sigma}{\sqrt n}

E = 1.960 * \frac{5}{\sqrt {150}}

E = 1.960 * \frac{5}{12.25}

E =  \frac{1.960 *5}{12.25}

E =  \frac{9.80}{12.25}

E =  0.800

Solving (d): The confidence interval

This is calculated as:

CI = (\bar x - E, \bar x + E)

CI = (7.10 - 0.800, 7.10 + 0.800)

CI = (6.3, 7.9)

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

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