Please help !<br><br> choose the definition for the function.

Find the composite area of the shaded region. Use 3.14 for 11.

Ganezh [65]

Answer:

7

Step-by-step explanation:

4 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

Can someone help me solve these two problems? i need to find what x is.

Bond [772]

Answer:

1. x=2

2. x= 0.33

Step-by-step explanation:

1.So to start off you have a coefficient for 2 so you have to find what 2 to the power of 2 is =4 so then you have your new problem 4x+5=13 next you need to get 4x by itself so

-5 -5

then you have 4x=8. you divide by 4 on both sides

4. 4

so then you get x=2

2.So same here find what 3 to the power of 2 is =9

New equation 4x-1=9x+4x-4. Then combine like terms on the right you cant do it on the left because that is a separate part and those dont merge like terms so

9x+4x = 13x

new equation. 4x-1=13x-4

next subtract 4x on both sides

13x-4x=9x

new equation. -1=9x-4

next add 4 to both sides

-1+4= 3

new equation. 3=9x

so then divide 9 on both sides

3÷9= 0.33

x= 0.33

Hope this made sense and helps

7 0

3 years ago

A is plotted on a coordinate grid at start bracket negative 3 and 1 over 2, 2 end bracket.. Choose the correct grid that best sh

ra1l [238]

Answer:

B

Step-by-step explanation:

have a good day and stay safe

5 0

3 years ago

Answer please its for my finals please

dalvyx [7]

Answer:

Part A: c = 2.25 + b

Part B: c = 6.75, b = 8

hope this helped!

5 0

2 years ago

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Please help ! choose the definition for the function.