Data Analysis - PSAT Math

The easiest way to solve this is to find the total number of miles traveled by the car and divide that by the total time travelled.

Recall that D = rt; therefore, for each of these three periods, we can calculate the distance and sum those products:

Dtotal = 60 * 3 + 55 * 2 + 40 * 3 = 180 + 110 + 120 = 410

The total amount of time travelled is: 3 + 2 + 3 = 8

Therefore, the average rate is 410 / 8 = 51.25 miles per hour.

Let's let a and b stand for the smallest and largest values in S, respectively. Because T is equal to the square of the elements in S, then the smallest and largest values in T will be equal to _a_2 and _b_2 , respectively.

We are told that the range of T is equal to 560. The range of a set of numbers is the difference between the largest and smallest, which in this case we can represent as _b_2 – _a_2. We can now write the following equation:

_b_2 – _a_2 = 560

Noice that _b_2 – _a_2 can be factored as (b – a)(b + a), which is the formula for the difference of squares.

(b – a)(b + a) = 560

Notice that the value b – a represents the range of S, because it is the difference between the largest and smallest values of S.

We are told that S consists of six consecutive odd integers. We can use this information to find the range of S. Because they are consecutive odd integers, each member in S will be two larger than the one before it. Thus, we could represent S as follows:

a, a + 2, a + 4, a + 6, a + 8, a + 10

The range of S is equal (a + 10) – a = 10. Thus, b – a must equal 10.

Let's go back to the equation (b – a)(b + a) = 560. We can replace b – a with 10 and solve for b + a.

(10)(b + a) = 560

Divide both sides by 10.

b + a = 56

So, we know that b – a = 10, and b + a = 56. We can solve this system of equations to find the values of a and b, which will give us the smallest and largest values in S.

Let's solve the first equation in terms of b.

b – a = 10

Add a to both sides.

b = 10 + a

Now we can substitute 10 + a in for b in the second equation.

(10 + a) + a = 56

10 + 2_a_ = 56

Subtract 10 from both sides.

2_a_ = 46

Divide both sides by 2.

a = 23

b = 10 + a = 10 + 23 = 33

The smallest value of S is 23, and the largest value is 33. Thus, S consists of the following set:

23, 25, 27, 29, 31, 33

We are told that T is equal to the squares of the elements of S. Thus, T consists of the following numbers:

529, 625, 729, 841, 961, 1089

We are asked ultimately to find the median of T. The median of any set is the middle number. In this case, T has six elements, so there are actually two numbers in the middle: 729 and 841. To find the median, we will take the average of the two middle numbers. The average of two numbers is equal to their sum divided by two.

median of T = (729 + 841)/2 = 1570/2 = 785.

The answer is 785.

The set consists of eight consecutive odd integers. Let's call the integers, in order from least to greatest, A, B, C, D, E, F, G, and H. We are told that the median of the set is 22. The median in a set of numbers is the number in the middle. In the set of A, B, C, D, E, F, G, and H, the numbers represented by D and E would be in the middle. When we have two members in the middle, the median is equal to their average. Thus, 22 is equal to the average of D and E. In other words, 22 is the average of the 4th and 5th numbers in the set. Since 22 is halfway between 21 and 23, it makes sense that the 4th and 5th numbers would have to be 21 and 23, respectively. Thus, our set looks like this:

A, B, C, 21, 23, F, G, H

Beacuse the set consists of consecutive odd integers, each integer is two larger than the one before it. Therefore, this is the set of integers:

15, 17, 19, 21, 23, 25, 27, 29

The problem asks us to find the largest integer in the set, which is 29.

The answer is 29.

To find the median, arrange the numbers from lowest to highest then find the middle number.

There are two numbers in the middle in this set ( and ).

In this case, the median is 6 but you would typically find the average of the two numbers in the middle.

The median is the "middle number" in a sorted list of number; therefore, reorder the numbers in numerical order: 7, 8, 12, 15, 19, 20. Since x is the middle number, it must come between 12 and 15, leaving 14 as the only viable choice.

The median is the coin which has an equal number of coins of lesser or equal value and greater or equal value on either side of it. The median therefore falls to one of the nickels (62 quarters, dimes and nickels above it and 62 nickels and pennies below it). The value of a nickel is 5 cents.

We first remove any 94%+ scores. For tests scored out of 50, a 94% is a 47. Thus anything 47 or greater will be removed. We see that this leaves us with:

For the second half, we have tests scored out of 40. 94% of 40 gives us 37.6, thus we will only remove tests that were a 38 or better. This leaves us with:

Now we have to convert them to percentiles. We end up with the lists as follows. The raw scores are in parentheses:

We then order it by percentile (removing raw scores, as they are unnecessary):

We notice that the median must be between the 78 and the 87.5 percentile scores. 31 out of 40 gives us 77.5%, which is below the score associated with an earlier test. Thus this cannot be the answer. The remaining 4 scores are all higher than 78%, and the largest of them is equal to 87.5 (the highest bound).

The median is the middle value in an ordered set of numbers. In this instance, we have 8 numbers, so the median is the average of the 4th and 5th number:

The median for this set is 3.5. Now let's find the mean, or the average, of the set:

Now that we have the mean and median, we can finish the question. The question asks for the mean (3.375) minus the median (3.5):

The mode of a problem is the most common number, i.e. the number that appears the most times in a given set of data. In this case, only one number is repeated. Therefore, the mode is 5.

The mode is the most frequently represented number in a data set. In this instance, that is and . There can be more than one mode.

To find the range, subtract the smallest number from the largest number. In this case we get 23 - 1 = 22 so the range is 22.

The range of a set of data is found by subtracting the smallest item from the largest. In this case, 843 - 103 = 740.

We don't know what the numbers are yet, so let's just call them a1, a2, a3, a4....a11. Let's assume a1 is the smallest, and a11 is the greatest.

If we were going to find the median of our set, we would need to line them all up, from least to greatest.

a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11

The median is the middle number, which in this case about be a6, because there are five numbers before it and five numbers after it.

We know that the median is 62, so that means that there must be five numbers before 62 and five numbers after. Because the numbers are all consecutive even integers, the set must look like this:

52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72

To find the range of this set, we must find the difference between the largest and smallest numbers. The largest is 72, and the smallest is 52, so the range is 72 - 52 = 20.

The answer is 20.

Let's represent the eleven integers as A, B, C, D, E, F, G, H, I, J, and K. Let's assume that if we lined them up from least to greatest, the set would be as follows:

A, B, C, D, E, F, G, H, I, J, K

The median in this set is the middle number. In this case, the middle number will always be F, because there are five numbers before it and five numbers after it.

We are told that the median is 39. Thus F = 39, and our set looks like this:

A, B, C, D, E, 39, G, H, I, J, K

Since we know that all the numbers are consecutive odd integers, the set must consist of the five odd numbers before 39 and the five odd numbers after 39. In other words, the set is this:

29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49

The question asks for the range of the set, which is the difference between the largest and smallest numbers. In this case, the range is 49 – 29 = 20.

The answer is 20.

First, arrange the numbers from highest to lowest, then subtract the highest number from the lowest number.

The range of a data set is the distance between the smallest number and the largest number. To find the range, you subtract the smallest number (5) from the largest number (13). Therefore, the range is 8.

To find the range of a data set, subtract the smallest number in the set from the largest:

The first thing to do here is find the total number of students in the contest. Seniors = 300 * 5 = 1500, Juniors = 200 * 3 = 600, Sophomores = 200, and Freshmen = 100. So adding all these up you get a total of 2400 names in the hat. Out of these 2400 names, 600 of them are Juniors. So the probability of choosing a Junior's name is 600/2400 = 1/4.

Here we have a logic statement:

If A (Jean's brother), then B (red hair).

"If Ron has red hair, then he is Jean's brother" states "If B, then A" - we do not know whether or not this is true.

"If Winston does not have red hair, then he is not Jean's brother" states "If not B, then not A" - this has to be true.

"If Paul is not Jean's brother, then he has red hair" states "If not A, then B." We do not know whether or not this is true.

"If Eddie is Jean's brother, then he does not have red hair" states "If A, then not B" - we know this cannot be true.