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2 years ago

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Mathematics

1 answer:

lesantik [10]2 years ago

6 0

Answer:

10x^2-11x-6

Step-by-step explanation:

To expand, we use the FOIL method. You do this by multiplying each number inside the parenthesis by the other numbers in the other set of the parenthesis. For example, on this problem, you would multiply 5x by 2x, and then 5x by -3.

Step by step, you would:

5x * 2x = 10x^2

5x * -3 = -15x

2 * 2x = 4x

2 * -3 = -6

In an equation, it would expand to 10x^2-15x+4x-6

To simplify, you add like terms (constants and those with the same variable). In this case, you would add together all the terms ending in x.

-15x+4x=-11x

Now, you have added like terms together, so your simplified answer is 10x^2-11x-6.

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Step-by-step explanation:

Step 1:

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7 0

2 years ago

[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

3 years ago