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Vanyuwa [196]
3 years ago
10

a set of date consists of 225 observations. the lowest value of the data set is 2,403; the highest is

Mathematics
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

8 classes

Step-by-step explanation:

Given

Least = 2403

Highest = 11998

n = 225

Required

The number of class

To calculate the number of class, the following must be true

2^k > n

Where k is the number of classes

So, we have:

2^k > 225

Take logarithm of both sides

\log(2^k) > \log(225)

Apply law of logarithm

k\log(2) > \log(225)

Divide both sides by log(2)

k > \frac{\log(225)}{\log(2)}

k > 7.8

Round up to get the least number of classes

k = 8

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Solve system by elimination <br> -5x-10y=-20<br> 10x+10y=0
bekas [8.4K]

Answer:

(-4,4)

Step-by-step explanation:

-5x-10y=-20

10x+10y=0

divide both sides by 2

-5x-10y=-20

5x+5y=0

that makes

-5y=-20

divide both sides by -5

y=4

now substitute 4 for y

10x+10 x 4=0

multiply

10x+40=0

subtract

10x=-40

divide

x=-4

7 0
3 years ago
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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
If the nth term of a sequence is 10n what is the 12th term
Lisa [10]

Answer:

10th term is 120.555

Step-by-step explanation:

6 0
3 years ago
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If LN = 6x - 5, LM = x + 7, and MN = 3x + 20, find MN.<br> A.68<br> B.16<br> C.23<br> D.18
borishaifa [10]

Answer : The correct option is (A) 68.

Step-by-step explanation :

As we are given that:

LN = 6x - 5

LM = x + 7

MN = 3x + 20

Now we have to determine the value of MN.

According to the question:

LN = LM + MN

Now putting all the given values in this expression, we get:

6x - 5 = (x + 7) + (3x + 20)

6x - 5 = 4x + 27

6x - 4x = 27 + 5

2x = 32

x = 16

The value of MN = 3x + 20 = 3(16) + 20 = 48 + 20 = 68

Therefore, the value of MN is 68.

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3 years ago
Which values are solutions to the inequality below? Check all that apply.
sp2606 [1]
2 I think that is the answer
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