Can somebody classify this figure

The data set below represents the ages of 36 executives. find the percentile that corresponds to an age of 4141 years old. 2828

solong [7]

Answer:

37th percentile.

Step-by-step explanation:

We have been given a data set that represents the ages of 36 executives. We are asked to find the percentile that corresponds to an age of 41 years.

28, 29, 29, 32, 32, 33, 34, 34, 34, 34, 37, 37, 38, 41, 41, 42, 45, 45, 47, 47, 47, 48, 50, 51, 53, 56, 56, 56, 61, 61, 62, 63, 64, 64, 65, 66.

Let us count the number of data points below and at 41.

We can see that the number of data points at and below 41 is 13.

We will use percentile formula to solve our given problem.

$\text{Percentile rank of x}=\frac{\text{Number of values below x}}{\text{Total number of data points}}\times 100$

$\text{Percentile rank of 41}=\frac{13}{36}\times 100$

$\text{Percentile rank of 41}=0.361111\times 100$

$\text{Percentile rank of 41}=36.11\approx 37$

Therefore, the percentile rank that corresponds to age of 41 years old is 37th percentile.

8 0

2 years ago

Emily folded a paper circle into 2 equal parts. What is the angle measure of each part?

timama [110]

The diameter splits the circle into 2 equal halves by going through the midpoint of the circle. The two halves are called semicircles.

7 0

2 years ago

Solve the inequality 47.75 + x Less-than-or-equal-to 50 to determine how much more weight can be added to Li's suitcase without

myrzilka [38]

Answer:

x is less-than-or-equal-to 2.25 (x ≤ 2.25)

Step-by-step explanation:

We can write down the inequality that represents the weight Li can add without going over the 50 pound limit:

47.75 + x ≤ 50

If we solve for x we have:

47.75 + x ≤ 50

x ≤ 50 - 47.75

x ≤ 2.25

Therefore, the weight Li can add to the suitcase is less-than-or-equal-to 2.25

6 0

2 years ago

Read 2 more answers

a lunch stand makes a $.75 profit on each chef's salad and $1.20 profit on each caesar salad. On a typical weekday, it sells bet

tangare [24]

Answer:

50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.

Step-by-step explanation:

Suppose, the number of Chef's salad is $x$ and the number of Caesar salad is $y$

On a typical weekday, it sells between 40 and 60 Chefs salads and between 35 and 50 Caesar salads.

So, the two constraints are: $40\leq x\leq 60$ and $35\leq y\leq 50$

The total number sold has never exceed 100 salads. So, another constraint will be: $x+y\leq 100$

According to the graph of the constraints, the vertices of the common shaded region are: $(40,35), (60,35), (60,40), (50,50)$ and $(40,50)$ (Refer to the attached image for the graph)

The lunch stand makes a $.75 profit on each Chef's salad and $1.20 profit on each Caesar salad. So, the profit function will be: $P=0.75x+1.20y$

For (40, 35) , $P=0.75(40)+1.20(35)=72$

For (60, 35) , $P=0.75(60)+1.20(35)=87$

For (60, 40) , $P=0.75(60)+1.20(40)=93$

For (50, 50) , $P=0.75(50)+1.20(50)=97.5$ (Maximum)

For (40, 50) , $P=0.75(40)+1.20(50)=90$

Profit will be maximum when $x=50$ and $y=50$

Thus, 50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.

6 0

2 years ago

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = $\{ (a,b) \in R^2 \}$

d) R1ºR4 = $\{ (a,b) \in R^2 \}$

e) R1ºR5 = R1

f) R1ºR6 = $\{ (a,b) \in R^2 \}$

g) R2ºR3 = $\{ (a,b) \in R^2 \}$

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that ac. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0

2 years ago