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I am Lyosha [343]
3 years ago
12

A circle has a diameter of 20 ft. What is the diameter? *

Mathematics
2 answers:
fenix001 [56]3 years ago
3 0

Answer:

20ft

Step-by-step explanation:

Montano1993 [528]3 years ago
3 0

Answer:

20 ft lol

Step-by-step explanation:

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Carla is going to draw a marble from the bag shown below, replace it, and then draw another marble. What is the probability that
Contact [7]

Answer:

7/80

Step-by-step explanation:

(3 green, 4 yellow, 5 blue, 8 pink) = 20 marbles

P ( green or yellow  ) = number of green or yellow / total

                = (3+4)/20  = 7/20

Replace  

so we have (3 green, 4 yellow, 5 blue, 8 pink) = 20 marbles

P (blue) = number of blue/ total

                = 5/20  = 1/4

P ( green or yellow, replace, blue) = 7/20 * 1/4 = 7/80          

8 0
3 years ago
The perimeter of the rectangle of the rectangle is 146 units. what is the length of the longer side ??? HELP ASAP
Novay_Z [31]

2x+2x+3x+3+3x+3=146

10x+6=146

10x=140

x=14

45 units is the longest side



Hope this helped!!!!!

6 0
3 years ago
Which pair of numbers has GCF of 5
finlep [7]
The only pairs for the greatest common factors of 5 are 1, 5
8 0
3 years ago
Find the slope of the line.<br><br> -4,1<br> 1,-2
const2013 [10]
Answer: finding slope (-4, 1), (1,-2) : m= -3/5
6 0
3 years ago
Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordl
LenaWriter [7]

Answer:

a) 0.0498

b) 0.1489

c) 0.1818

Step-by-step explanation:

Given:

Number of telephones = 6+6+6= 18

6 cellular, 6 cordless, and 6 corded.

a) Probability that all the cordless phones are among the first twelve to be serviced:

12 are selected from 18 telephones, possible number of ways of selection = ¹⁸C₁₂

Then 6 cordless telephones are serviced, the remaining telephones are: 12 - 6 = 6.

The possible ways of selecting thr remaining 6 telephones = ¹²C₆

Probability of servicing all cordless phones among the first twelve:

= (⁶C₆) (⁶C₁₂) / (¹⁸C₁₂)

= \frac{1 * 924}{18564}

= 0.0498

b) Probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced:

Here,

One type must be serviced first

The 6 remaining to be serviced can be a combination of the remaining two types.

Since there a 3 ways to select one type to be serviced, the probability will be:

= 3 [(⁶C₁)(⁶C₅) + (⁶C₂)(⁶C₄) + (⁶C₃)(⁶C₃) + (⁶C₄)(⁶C₂) + (⁶C₅)(⁶C₁)] / ¹⁸C₁₂

= \frac{3 * [(6)(6) + (15)(15) + (20)(20) + (15)(15) + (6)(6)]}{18564}

= \frac{2766}{18564}

= 0.1489

c) probability that two phones of each type are among the first six:

(⁶C₂)³/¹⁸C₆

\frac{3375}{18564}

=0.1818

5 0
4 years ago
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