Two standardized tests, a and b, use very different scales of scores. the formula upper a equals 40 times upper b plus 50a=40×
b+50 approximates the relationship between scores on the two tests. use the summary statistics for a sample of students who took test b to determine the summary statistics for equivalent scores on test a. lowest score equals= 2121 mean equals= 2929 standard deviation equals= 22 q3 equals= 2828 median equals= 2626 iqr equals= 66 find the summary statistics for equivalent scores on test a. lowest scoreequals= nothing meanequals= nothing standard deviationequals= nothing q3equals= nothing medianequals= nothing iqrequals= nothing
Adding (or subtracting) a constant to every data value adds (or subtracts) the same constant to measures of position such as center,percentiles, max or min.
Its shape and spread such as range, IQR, standard deviation remain unchanged. When we multiply (or divide) all the data values by any constant, all measures of position (such as the mean, median, and percentiles) and measures of spread (such as the range, the IQR, and the standard deviation) are multiplied (or divided) by that same constant. Part A:
The lowest score is a measure of location, so both addition and multiplying the lowest score of test B by 40 and adding 50 to the result will affect the lowest score of test A.
Thus, the lowest score of test A is given by 40(21) + 50 = 890
Therefore, the lowest score of test A is 890.
Part B:
The mean score is a measure of location, so both
addition and multiplying the mean score of test B by 40 and adding 50
to the result will affect the lowest score of test A.
Thus, the mean score of test A is given by 40(29) + 50 = 1,210
Therefore, the mean score of test A is 890.
Part C:
The standard deviation is a measure of spread, so multiplying the standard deviation of test B by 40 will affect the standard deviation but adding 50
to the result will not affect the standard deviation of test A.
Thus, the standard deviation of test A is given by 40(2) = 80
Therefore, the standard deviation of test A is 80.
Part D
The Q3 score is a measure of location, so both
addition and multiplying the Q3 score of test B by 40 and adding 50
to the result will affect the Q3 score of test A.
Thus, the Q3 score of test A is given by 40(28) + 50 = 1,170
Therefore, the Q3 score of test a is 1,170.
Part E:
The median score is a measure of location, so both
addition and multiplying the median score of test B by 40 and adding 50
to the result will affect the median score of test A.
Thus, the median score of test A is given by 40(26) + 50 = 1,090
Therefore, the median score of test A is 1,090.
Part F:
The IQR is a measure of spread, so multiplying the IQR of test B by 40 will affect the IQR but adding 50
to the result will not affect the IQR of test A.
The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be . The magnitude of will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is , we have:
(Recall that )
Now that we've found the vertical component of the velocity and launch, we can use kinematics equation to solve this problem, where is final and initial velocity, respectively, is acceleration, and is distance travelled. The only acceleration is acceleration due to gravity, which is approximately . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.
The dotted lines are parallel because they have the same <u>gradient</u> to each other, this means the two dotted lines will <u>never</u> touch each other.
