A study carried out to investigate the distribution of total braking time (reaction time plus accelerator-to-break movement time, in ms) during real driving conditions at 60 km/hr gave the following summary information on the distribution of times ("A Field Study on Braking Responses Dring Driving," Ergonomics, 1995: 1903-1910): mean = 535 , s = 80 . Assume the braking times have approximately a bell shaped distribution. A) Approximately what percentage of braking times are in the interval (535, 615)?B) What value do the top 2.5% of braking times exceed? (show work)

a) Approximately 34% of braking times are in this interval.

b) 695 ms

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 535

Standard deviation = 80

A) Approximately what percentage of braking times are in the interval (535, 615)?

The empirical rule is symmetric, which means that 50% of the measures are below the mean and 50% are above.

615 = 535 + 80

So 615 is one standard deviation above the mean.

535 is the mean.

Of the measures above the mean(50%), 68% are within 1 standard deviation of the mean.

Then

0.5*0.68 = 0.34

Approximately 34% of braking times are in this interval.

B) What value do the top 2.5% of braking times exceed?

95% of the measures are within 2 standard deviation of the mean. That is, between the bottom 100 - (95/2) = 2.5% and the top 100 + (95/2) = 2.5% are within 2 standard deviations of the mean.

The top 2.5% are more than two standard deviations of the mean.

535 + 2*80 = 695

So the top 2.5% of braking times exceed 695 ms.

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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