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Olegator [25]
3 years ago
6

A boat is moving at a velocity of –14 miles per hour.

Mathematics
1 answer:
Rama09 [41]3 years ago
5 0

Answer:

d

Step-by-step explanation:

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The number 42 is what percent of 35
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35 - 100%

42 - x


X=42×100/35=120%


120-100=20%


Answer: 20%

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What is the first step in solving the equation by factoring?
Art [367]

The answer is D. This is because all of the variables must be on one side so it can be factored.


8 0
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How do you find slope intercept from a graph
Makovka662 [10]

The slope intercept is equal to the rise/run.


So lets say to get from one point to the other, you go down 2 units and go right 3 units.


Since you went down it is a -2. If you had gone up 2 units, then it would be a positive 2.

Since you went to the right, it is a positive 3. If you had gone left 3 units, then it would be -3.


-2/3

4 0
3 years ago
Required information Skip to question NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to re
pogonyaev

The result for the given 14 length bit string is-

(a) The bit strings of length 14 which have exactly three 0s is 2184.

(b) The bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The bit strings of length 14 which have at least three 1s is 4472832.

<h3>What is a bit strings?</h3>

A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.

A null string is a bit-string with no length.

The concept used here is permutation;

ⁿPₓ = n!/(n-x)!

Where, n is the total samples.

x is the selected samples.

(a) Because there are exactly three 0s, its other digits have always been one, hence the total of permutations is equal to;

n = 14 and x = 3 digits.

¹⁴P₃ = 14!/(14-3)!

¹⁴P₃ = 2184.

Thus, the bit strings of length 14 which have exactly three 0s is 2184.

(b) Because it is a bit string with 14 digits and it can only have digits 0 or 1, we must choose 7 0s and the remaining 7 1s, hence the number possible permutations is equal to;

n = 14 and x = 7 digits.

¹⁴P₇ = 14!/(14-7)!

¹⁴P₇ = 17297280.

Thus, the bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The answer is identical to problem a, but the rest of a digits could be either 0 or 1, hence it must be doubled by 2¹¹ because there are 11 digits, each of which can be one of two options;

= 2184×2¹¹

= 4472832

Thus, he bit strings of length 14 which have at least three 1s is 4472832.

To know more about permutation, here

brainly.com/question/12468032

#SPJ4

The correct question is-

How many bit strings of length 14 have:

(a) Exactly three 0s?

(b) The same number of 0s as 1s?

(d) At least three 1s?

3 0
1 year ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!<br><br> Factor.<br><br> 11x^2 + 35x + 6
fredd [130]

Hey there!!

Given equation :

11x² + 35x + 6

Now let's write 35x as 33x and 2x

Then the equation would become :

... 11x² + 33x + 2x + 6

... Now, let's take the common terms 11x² + 33x and 2x + 6

It would become :

... 11x ( x + 3 ) + 2 ( x + 3 )

... Now, we will write this as :

... ( 11x + 2 ) ( x + 3 )

Hence, this is as the answer...

Hope my answer helps!!

6 0
2 years ago
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