Which of the following is a counterexample for this conditional statement?

[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6

viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

Height (in inches) of a 25 year old man is a normal random variable with mean $g=71$ and variance $o^{2} =6.25$ .

To find: (a) What percentage of 25 year old men are 6 feet, 2 inches tall

(b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X> $74in$ ]

P[X>74] = P[ $\frac{X-g}{o}$ > $\frac{74-71}{2.5}$ ]

= P[Z > 1.2]

= 1 - P[Z ≤ 1.2]

= 1 - Ф (1.2)

= 1 - 0.8849

= 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, $6ft$ $5in$ - X, $6ft$ ]

P[X > $6ft$ $5in$ I X > $6ft$ ] = P[X > 77 I X > 72]

= $\frac{P[X > 77]}{P[ X > 72]}$

= $\frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}] }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }$

= $\frac{P[Z >2.4]}{P[Z>0.4]}$

= $\frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }$

= $\frac{1-0.9918}{1-0.6554}$

= $\frac{0.0082}{0.3446}$

= 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0

2 years ago

___ cosb =1/2 sin(a+b)+sin(a-b)?

vodomira [7]

Answer:

That would be sina.

Step-by-step explanation:

sin(a+b) = sinacosb + cosasinb

sin(a-b) = sinacosb - cosasinb

Adding we get sin(a+b) + sin(a-b) = 2sinaccosb

so sinacosb = 1/2sin(a+b) + sin(a-b)

8 0

1 year ago

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HELP 100 POINTS GIVEN 1. An inground non-diving swimming pool has the following dimensions. It is 33 feet long and 14 feet wide.

olga2289 [7]

Answer:

14inches farenheight

Step-by-step explanation:

3 0

2 years ago

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walnut grower estimates from past records that if 20 trees are planted per acre, then each tree will average 60 pounds of nuts p

laila [671]

Answer:

5 trees should be planted to maximize the yield per acre,

The maximum yield would be 1250

Step-by-step explanation:

Given,

The original number of trees per acre = 20,

Average pounds of nuts by a tree = 60,

Let x be the times of increment in number of trees,

So, the new number of trees planted per acre = 20 + x

∵ for each additional tree planted per acre, the average yield per tree drops 2 pounds,

So, the new number of pounds of nut = (60 - 2x)

Thus, the total yield per acre,

$Y(x) = (20+x)(60-2x)$

Differentiating with respect to t ( time ),

$Y'(x) = (20+x)(-2) + 60 - 2x = -40 - 2x + 60 - 2x = 20 - 4x$

Again differentiating with respect to t,

$Y''(x) = -4$

For maxima or minima,

$Y'(x) = 0$

⇒ 20 - 4x = 0

⇒ 20 = 4x

⇒ x = 5,

For x = 5, Y''(x) = negative,

Hence, Y(x) is maximum for x = 5,

And, maximum value of Y(x) = (20+5)(60 - 10) = 25(50) = 1250,

i.e. 5 trees should be planted to maximize the yield per acre,

and the maximum yield would be 1250 pounds

4 0

1 year ago

The graph represents the piecewise function: f(x) = What is the domain and range of the function? Domain: Range:

White raven [17]

Answer:

domain: All real numbers

Range: all real numbers greater than or equal to 0

Step-by-step explanation:

Edge 2020

4 0

1 year ago

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