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Alborosie
3 years ago
12

Please help me with questions 14 and 15

Mathematics
1 answer:
Likurg_2 [28]3 years ago
8 0
Problem 14: The answer is choice A since the x coordinate flips in sigh (from positive to negative or vice versa). The y coordinate stays the same. This rule reflects the figure over the y axis.

---------------------------------------------------------------------------------------------

Problem 15: The answer is choice A
cos(angle) = adjacent/hypotenuse
cos(B) = BC/AB
cos(B) = 12/13
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Slope intercept from 2x+2y=10
vampirchik [111]

y=x-5

Thatll be one brainliest

4 0
3 years ago
If 35% of a number is 21 and 90% of the same number is 54, find 55% of that number.
inn [45]

Answer:

Step-by-step explanation:

The number is 60:

Divide the amount and decimal of the percentage to be extracted

Case 1

\frac{21}{0.35} =60

Case 2

\frac{54}{0.9} =60

Now the 55% of this number is:

60(0.55)=33

Hope this helps

4 0
2 years ago
Round 678,483 to the nearest hundred thousand
White raven [17]
700,000 is the answer...i think
8 0
3 years ago
Read 2 more answers
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
Guess what it says for brainliest
bulgar [2K]

Answer:

Let's get this show on the road!

Step-by-step explanation:

I'm not sure how to explain it, i was just thinking of phrases that fit the spaces.

3 0
3 years ago
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