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wlad13 [49]
3 years ago
12

How much space is given for each parking spot at the mall parking lot if each spot has an equal width?

Mathematics
1 answer:
choli [55]3 years ago
7 0

Answer: 3.13 m

Step-by-step explanation:

There are 10 parking spots at the mall.

The combined width of those parking spots is 31.30m

If all 10 of the parking spots have equal width, the width of each spot is:

= 31.30 / 10 parking spots

= 3.13 m

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Andre45 [30]
(5,4)

On a system of 2 perpendicular axis with O as origine, the pair (5,4) means:

5 is the distance from the origin O and situated on x-axis (on the right of O)
4 is the distance from the origin O and situated on y-axis (above O)

Then the pair (5,4) is situated in the 1st Quadrant

8 0
3 years ago
Your savings account pays 2.5% interest per year. How much interest does the account pay per month.
Mkey [24]

Answer:

0.21%

Step-by-step explanation:

2.5/12=0.208 or 0.21

3 0
3 years ago
A 30-foot ladder is leaning against an office building. if the base of the ladder is
Lina20 [59]

Angle of elevation would be 66 degrees

4 0
2 years ago
A pharmacist receives a shipment of 22 bottles of a drug and has 3 of the bottles tested. If 5 of the 22 bottles are contaminate
Norma-Jean [14]

Answer:

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.

Step-by-step explanation:

C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this problem, we have that:

Total number of combinations:

C_{22,3} = \frac{22!}{3!(18)!} = 1540

Desired combinations:

It is 1 one 5(contamined) and 2 of 17(non contamined). So:

C_{5,1}*C_{17,2} = 5*17*8 = 680

What is the probability that exactly 1 of the tested bottles is contaminated?

P = \frac{680}{1540} = 0.4416

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.

5 0
3 years ago
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