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Andrej [43]
3 years ago
11

Can someone plz help me with this? Which polygons are similar?

Mathematics
1 answer:
earnstyle [38]3 years ago
7 0
The first one I believe!
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Find the perimeter of each rectangle.
alexdok [17]

Answer: A) 66 In

Step-by-step explanation:

2 sides are 23 Inches

2 Sides are 10 Inches

So you add 23 + 23 + 10 + 10 = 66 to get the perimeter

4 0
3 years ago
Mily wrote the equation −3x=−2x−6
Dima020 [189]

Answer:

x=6

Step-by-step explanation:

-3x=-2x-6

-3x+2x=-6

-x=-6

x=6

4 0
3 years ago
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I’ll brainliest you please help me
Tanya [424]
X= 75
y= 105
this is because x is equal to the angle with 75, y us 105 because 180-75=105. 180 is found because the angle of a line is 180
3 0
3 years ago
Solve for x 5x + 3 - 2x = 24 (also don't have the answer be a fraction)
Yuki888 [10]

Answer:

x=7

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+3−2x=24

Step 1: Simplify both sides of the equation.

5x+3−2x=24

5x+3+−2x=24

(5x+−2x)+(3)=24(Combine Like Terms)

3x+3=24

3x+3=24

Step 2: Subtract 3 from both sides.

3x+3−3=24−3

3x=21

Step 3: Divide both sides by 3.

7 0
3 years ago
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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
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