A quadrilateral with two pairs of parallel sides and no right angles

The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.

koban [17]

The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

Revenue is Q₂ < $7,500 < Q₃

The sales for the week is better than 72.91% of all sales

The performance for the week for Bridge Road

Revenue; Q₂ < $7,100 < Q₃

The sale for the week is better than 59.87% of all sales

(b) The mean is $3611

The median is $3,600

The standard deviation is $3250

The Interquartile range is $6075

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

$\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right]$

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

$Z = \dfrac{x - \mu }{\sigma }$

$Z = \dfrac{7100 - 5,989 }{1794 } \approx 0.6193$

From the Z-Table, we have;

The percentile= 0.7291

Therefore, the sale for the week for Park Street is better than 72.91% of all the sales

Park Street revenue;

The z-score is given as follows;

$Z = \dfrac{7500 - 6611}{3580} \approx 0.25$

From the Z-Table, we have;

The percentile = 0.5987

Therefore, the sale for the week is better than 59.87 % of all the sales

(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

The new mean = $3,611

The new median = 6600 - 3000 = 3600

The new median = $3,600

The standard deviation and the interquartile range remain the same, therefore, we have;

The standard deviation = $3,580

The interquartile range = 9675 - 3600 = 6075

The interquartile range = 6075

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5 0

2 years ago

lauren planted 20 flowers, but her neighbors dog ate 7 of them. what percent of the flowers did the dog eat

LuckyWell [14K]

35 percent got eaten.

To solve: you’d take the number of flowers eaten, 7. And you’d divide it by the overall number, 20. Then you’d get 0.35, which would be your percent.

8 0

3 years ago

Find the midpoint of the segment whose endpoints are (-6,-8) and (3,6) show your work!

pogonyaev

Answer:

-1.5, -1

Step-by-step explanation:

midpoint formula

(x1 + x2 / 2) , (y1 + y2/ 2)

-6 + 3 / 2 , -8 + 6 / 2

3 0

3 years ago

Read 2 more answers

PLEASE HELP ME WHAT AM I SUPPOSE TO DO?!

stepladder [879]

Answer:

Step-by-step explanation:

Angle 2= 50

angle 1= 180-50= 130 degrees (linear pair (sum 180))

angle 3= 130 degrees (vertically oposite angles are equal)

angle 6= 130 degrees (alternate interior angles)

angle 5= 180-130= 50 degrees (linear pair)

i hope this helps :)

8 0

3 years ago

A bag contains 8 purple marbles & 12 green marbles. You choose one marble out of the bag without looking.

MrRissso [65]

Answer:

1. P(P) = 8/20 = 0.4

2. P(G) = 12/20 = 0.6

Step-by-step explanation:

Given;

Number of green marbles G = 12

Number of purple marbles P = 8

Total T = 12+8 = 20

The probability that you choose a purple marble P(P) is;

P(P) = number of purple marbles/total number of marbles

P(P) = P/T = 8/20 = 0.4

P(P) = 0.4

The probability that you choose a Green marble P(G) is;

P(G) = number of Green marbles/total number of marbles

P(G) = G/T = 12/20 = 0.6

P(G) = 0.6

6 0

3 years ago