Is p=6m+9m an example of a proportional relationship?

heyy soo i just started math class and im veryy confused I've tried different things to get to the answer but it didn't work soo

german

yes sure, i learned this before so

1. a. make a proportion here, of percentage. $\frac{225}{250} =\frac{x}{100}$ if you cross multiply and divide you get 90%.

1. b. obviously the empty seats would be 10%.

2. a. proportion again, $\frac{0.90}{6.00} =\frac{x}{100}$ cross multiply then divide you ge 15%

3. 30% in decimal form is 0.3. multiply it then to the total amount, so 0.3 x 15 = 4.5

7 0

3 years ago

Read 2 more answers

3(x 2) 11 is greater than 20

Allushta [10]

3(x+2)+11>20
minus 11 both sides
3(x+2)>9
divide both sides by 3
x+2>3
minus 2 both sides
x>1

4 0

2 years ago

Which expression is equivalent to -6x^3 • x • y^2 • y^6?

Georgia [21]

Use the Product Rule: x^ax^b = x^a + b

6x^3 + 1y^2 + 6

Simplify 3 + 1 to 4

6x^4y^2 + 6

Simplify 2 + 6 to 8

4 0

3 years ago

Find the missing length. The triangles are similar.

Marina CMI [18]

Answer:

RK= 13 SK= 12

Step-by-step explanation

First, we find the missing length of RK using the given information: 65-52=13

Then we divide 65 by 13 to get a scale factor: 65/13=5

(To check, we multipuly: 13*5=65)

We then apply the scale factor to MK to find SK: 60/5=12

(To check we multipuly: 5*12=60)

We now have our answers: RK:13, SK:12

4 0

2 years ago

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

2 years ago