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Ad libitum [116K]
3 years ago
9

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

Mathematics
1 answer:
kolezko [41]3 years ago
4 0
A Parallelogram. 2 obtuse angles, 2 acute angles. 2 pairs of parallel sides
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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 <u>Q₂ < $7,500 < Q₃</u>
  • The sales for the week is better than <u>72.91%</u> of all sales

The performance for the week for Bridge Road

  • Revenue; <u>Q₂ < $7,100 < Q₃</u>
  • The sale for the week is better than <u>59.87%</u> of all sales

(b) The mean is <u>$3611</u>

The median is $<u>3,600</u>

The standard deviation is $<u>3250</u>

The Interquartile range is $<u>6075</u>

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 <u>72.91%</u> 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 = <u>0.5987</u>

  • Therefore, the sale for the week is better than <u>59.87 %</u> 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 = <u>$3,611</u>

The new median = 6600 - 3000 = 3600

  • The new median = <u>$3,600</u>

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

  • The standard deviation = <u>$3,580</u>

The interquartile range = 9675 - 3600 = 6075

  • The interquartile range = <u>6075</u>

Learn more here:

brainly.com/question/21133077

brainly.com/question/23305909

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
2 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 &amp; 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
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