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Ilya [14]
3 years ago
11

Similar triangles. Solve for x

Mathematics
2 answers:
Vanyuwa [196]3 years ago
8 0
X would equal 3 i am just guessing so its not 100%
levacccp [35]3 years ago
7 0
(11x-4) / 60 = 70/50
550x -200 = 4200
550x = 4400
x = 4400 / 550
x =8
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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8
Delvig [45]
We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34
Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86

Thus the 98% confidence interval will be (105.34, 141.86)


Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
3 0
3 years ago
50×60×8000 f(x)=4000× f (y)=​
Aloiza [94]

Answer:

9,600,000,000 F (y)=5000

Step-by-step explanation:

hope for help po

4 0
3 years ago
Evaluate x + 23<br> when x=7
iren2701 [21]

Answer:

30

Step-by-step explanation:

x + 23 = ?

x = 7

All we need to do is plug in x =7 to the first equation:

7 + 23 = ?

We simplify and get:

30 = ?

3 0
2 years ago
The height of a triangle is 8cm more than the base. If the area is 90cm^2, find the base and the height
Aleonysh [2.5K]
I get the base to be 4.74 and the height to be 12.74
3 0
3 years ago
What is the side length of a square with the area of 20cm2
rodikova [14]

Answer:

S = a² => 20 = a² => a = √20 ≈ 4.5 cm

Step-by-step explanation:

5 0
3 years ago
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