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liraira [26]
2 years ago
8

How much be the value of x in y_axix?​

Mathematics
1 answer:
nikklg [1K]2 years ago
7 0

Answer:

i hope it will be help you

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Plz help and thanks so much for it
kirza4 [7]
B+c+(c-b)+(b-c). That is your expression.  b 11 c 16
       
          First, let's convert the variables to real numbers:   11+16+(16-11)+(11-16)
             Now, let's solve that equation.                                    11+16+5+(-5)
                      5+(-5) cancels out, so all we have left is:                   27
                That is your perimeter.
    
7 0
3 years ago
Select all of the solutions to the system graphed below.​
ra1l [238]

0,-2

4,4

2,0

3,5

Step-by-step explanation:

that was my answer

3 0
3 years ago
How do i make a flowchart for a robot taking someone’s food order
Margarita [4]

Answer:

you basically make boxes and have arrows connecting them, then you'll add the information on them.

Step-by-step explanation:

it's basically like a pedigree table but less complicated. if you need help on what a flowchart looks like just go to google

5 0
3 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
Find the area and perimeter of the shape above
ELEN [110]

Answer:

Perimeter: 18.5

Area: 15

Step-by-step explanation:

P = a + b + c

P = 5 + 6 + 7.5

P= 18.5

A = 1/2bh

A = 1/2*7.5*4

A= 15

5 0
3 years ago
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