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Whitepunk [10]
1 year ago
11

Pls look at pic!!!!!!!!!!!!

Mathematics
1 answer:
Alekssandra [29.7K]1 year ago
8 0

Answer:

I don't think the answer is in the options.

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What is the value of the expression 3a+b2 when a=14 and b=32 ?
soldi70 [24.7K]

Answer:

106

Explanation:

3a+b2

Substitute the values for the variables:

3(14) + 2(32)

Solve:

3(14) + 2(32)

42 + 64

106

5 0
2 years ago
I will love and rate 5.0 if done correctly with no images and no trolling for answer.
Murljashka [212]
The correct answer should 4% because the ratio is 3 out of 75 which would leave 72 non-defective bulbs. So your answer is 4%
5 0
2 years ago
10.
vodomira [7]
<h3>Solution:</h3>
  • Given, side (a) = x+3
  • We know, surface area of a cube = 6a^2
  • Therefore, the surface area of the cube
<h3> </h3><h3>6(x + 3) ^{2}  \\  = 6( {x}^{2}  + 6x + 9) \\  = 6 {x}^{2}  + 36x + 54</h3><h3>Answer:</h3>

The answer is

{6x}^{2}  + 36x + 54

<h3>Hope it helps.</h3>

Sorry for delay

3 0
2 years ago
Of the 9-letter passwords formed by rearranging the letters AAAABBCCC (4 A's, 2 B's, and 3 C's), I select one at random. Determi
Tanya [424]

Answer:

a) 3

b) (8!/9!)-(7!/9!)

c) (1-(8!/9!))*(7!/9!)

Step-by-step explanation:

a)With 4 As ;  2Bs and 3Cs it is possible to get a palindrome if you fixed the  letters C according to: (2) in the extremes of the word and the other one at the center therefore you only have palindrome in the following cases

<u>C</u> (       ) <u>C</u> (       ) <u>C</u>

To fill in the gaps we have  4 letters  A and 2 letters B, wich we have two divide in two palindrome gaps,  

AAB         and    BAA the palindrome is  C  AAB C BAA C

BAA         and    AAB    "           "           is  C  BAA C AAB C  

ABA         and    ABA    "           "           is  C  ABA C ABA C

b) 4 A  ;   2B  ; 3C

We have the total number of elements  9, so the total number of possible outcomes is : 9!

Total events: 9!

if we fixed 3 C we have (the group of 3 Cs becoming one element) so the total amount of events with 3 adjacent Cs is: 7!

Therefore the probability of having 3 adjacent Cs is: 7!/9!

If we fixed only 2 Cs we have:

4 A  ; 2 B  ; 2C  : 1C

Total number of words (events) in this case is 8! (2C becomes 1 element)

so the total numbers of events is 8! the probability in this case is 8!/9!(this value includes cases of adjacent 3 Cs previous calculated ) so this value minus the case of 3 adjacent Cs ) give us 2 adjacent C and the other no next to them

Probability (of words with 2 adjacent Cs and the other no next to them is); 8!/9! - 7!/9!

c) Probability of B apart from each other is the whole set of events minus those where 2 B are adjacent or (become 1 element)

4 A ; 2B ; 3C

Total of events 9! and events with adjacent B is 8!/9!

Therefore the probability of words with 3 adjacent Cs and 2 B separeted is

the probability of 3 adjacent Cs (7!/9!) times probability of words with no adjacent Bs wich is (1-(8!/9!))*(7!/9!)

5 0
3 years ago
1/3 x (21.69 – 24.99)
OlgaM077 [116]

Answer:

I think it might be 0.30

Step-by-step explanation:

5 0
2 years ago
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