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3 years ago

Which function below represents the arithmetic sequence 3, 7, 11, 15...?

2 answers:

4 0

D) f(n) = 3 + 4(n-1)

3 = 1st term
4 = common difference among the terms
n = term number you are looking for.

To check: 3, 7, 11, 15, ...

f(1) = 3 + 4(1-1) = 3 + 4(0) = 3 + 0 = 3
f(2) = 3 + 4(2-1) = 3 + 4(1) = 3 + 4 = 7
f(3) = 3 + 4(3-1) = 3 + 4(2) = 3 + 8 = 11
f(4) = 3 + 4(4-1) = 3 + 4(3) = 3 + 12 = 15

never [62]3 years ago

4 0

Answer:

its d

Step-by-step explanation:

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Help!!!! <br> What is f(-5) if f(x) = |2x – 1] + 10?

quester [9]

It is 1
Explanation
[2(-5)-1]+10
[-10-1]+10
-11+10
=1

3 0

3 years ago

Can somebody help me?

Degger [83]

A percentage
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5 0

3 years ago

Suppose a normal distribution has a mean of 62 and a standard deviation of

lutik1710 [3]

Answer:

62.5%

Step-by-step explanation:

(56 - 62) / 4 = -1.5 z score = 0.0668

(64 - 62) / 4 = 0.5 z score = 0.6915

0.6915 - 0.0668 = 0.6247

Rounded this becomes 62.5%

7 0

2 years ago

Find the vertices and foci of the hyperbola with equation quantity x plus 1 squared divided by 225 minus the quantity of y plus

Lera25 [3.4K]

Answer:

The vertices are (-16,-5) and (14,-5)

The foci are (-26,-5) and (24,-5)

Please, see the attached file.

Thanks.

7 0

3 years ago

Read 2 more answers

Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m

strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0

3 years ago