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andreyandreev [35.5K]
3 years ago
14

Heights of men have a bell-shaped distribution, with a mean of 176 cm and a standard deviation of 7 cm. Using the Empirical Rule

, answer the following questions: a) What is the approximate percentage of men between 169 and 183 cm? b) Between which 2 heights would 95% of men fall? c) Is it unusual for a man to be more than 197 cm tall? Explain.
Mathematics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

a) 68% of the men fall between 169 cm and 183 cm of height.

b) 95% of the men will fall between 162 cm and 190 cm.

c) It is unusual for a man to be more than 197 cm tall.

Step-by-step explanation:

The 68-95-99.5 empirical rule can be used to solve this problem.

This values correspond to the percentage of data that falls within in a band around the mean with two, four and six standard deviations of width.

<em>a) What is the approximate percentage of men between 169 and 183 cm? </em>

To calculate this in an empirical way, we compare the values of this interval with the mean and the standard deviation and can be seen that this interval is one-standard deviation around the mean:

\mu-\sigma=176-7=169\\\mu+\sigma=176+7=183

Empirically, for bell-shaped distributions and approximately normal, it can be said that 68% of the men fall between 169 cm and 183 cm of height.

<em>b) Between which 2 heights would 95% of men fall?</em>

This corresponds to ±2 standard deviations off the mean.

\mu-2\sigma=176-2*7=162\\\\\mu+2\sigma=176+2*7=190

95% of the men will fall between 162 cm and 190 cm.

<em>c) Is it unusual for a man to be more than 197 cm tall?</em>

The number of standard deviations of distance from the mean is

n=(197-176)/7=3

The percentage that lies outside 3 sigmas is 0.5%, so only 0.25% is expected to be 197 cm.

It can be said that is unusual for a man to be more than 197 cm tall.

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Which expression has a value of 66? A. (16 ÷ 3.2) × 2 + 0.2 × (8 + 82) B. [(16 ÷ 3.2) × (2 + 0.2)] × 8 + 82 C. 16 ÷ [(3.2 × 2) +
Setler [38]

Answer: none

Step-by-step explanation:

(A)

(16÷32/10) ×2 + 0.2×(90)

Using bodmas principle ; solve bracket

(16×10/32)×2 + (2/10×90)

10+18 =28

(B)

{(16÷32/10) × (2+2/10)} ×90

Open brackets

{(16×10/32) × (22/10)} ×90

(5×11/5) ×90

11×90 = 990

(C)

16÷{(32/10×2) + (2/10×8)} +82

Open brackets, solve division first, dolled by addition

16÷(32/5 + 8/5) +82

16÷(40/5) +82

16÷8 +82

2+82= 84

(D)

[16÷(32/10 ×2) + 0.2× (90)]

16÷ (32/5) + 2/10 ×90

Solve division

16×5/32 + 18

5/2 + 18

L.c.m of denominator (2&1) =2

(5+36) / 2 = 41/2

=20.5

8 0
2 years ago
When a number is divided by 5, it has a remainder of 4.
Alexxx [7]

Answer:

37

Step-by-step explanation:

I plugged in the numbers between 35 and 40 and got this. No problem. :)

7 0
2 years ago
A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The
disa [49]

Answer:

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Step-by-step explanation:

The null and alternative hypothesis are given by

H0: σ₁²= σ₂² against Ha: σ₁² ≠ σ₂²

Confidence interval for the population mean difference is given by

(x`1- x`2) ± t √S²(1/n1 + 1/n2)

Where S ²= (n1-1)S₁² + S²₂(n2-1)/n1+n2-2

Critical value of t with n1+n2-2= 50+ 35-2= 83 will be -1.633

Now calculating

S ²=34* (12.8)²+ (14.6)²*49/83= 192.96

Now putting the values in the t- test

(75.1 -72.1) ± 1.633 √ 192.96(1/35 +1/50)

=3 ±  5.09

=-2.09, 8.09  is the 90 % confidence interval for the difference

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

5 0
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lidiya [134]
6(7)^2+4(7)+8=y
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7 0
3 years ago
Read 2 more answers
20 points and will give brainliest!!
FromTheMoon [43]

Answer:

C=0.65M+22.65

Step-by-step explanation:

The standard charge includes an initial fee and plus an addditional fee for each mile driven and is given by:

S=0.40M+17.75

The insurance charge is given by:

I=0.25M+4.90

So, the total charge C will be the sum of the standard charge S and the insurance charge I:

C=S+I

Substitute:

C=(0.40M+17.75)+(0.25M+4.90)

Rearrange:

C=(0.40M+0.25M)+(17.75+4.90)

Add:

C=0.65M+22.65

And we have acauired our equation relating C to M!

4 0
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