15% of the skittles in andys bag are green. how many skittles are in the bag if 9 of them are green

Answer:

21 I think, might be wrong.

Answer:

no rocky

Step-by-step explanation:

Answer:

The answer is C

Step-by-step explanation:

I think it's C and I am always right so it is C! You're welcome!

Answer:

a) % age of samples containing none of these defects = 93.9%

b)% age of samples containing at least one of these defects = 6.1%

c) % age of samples free of type X and/or type Y defects = 95.8%

d) %age of samples with not more than 1 defect = 98.9%

Step-by-step explanation:

Data Given:

Number of Samples = 1000

Type X defect = 2.1% = 21 samples

Type Y defect = 2.4% = 24 samples

Type Z defect = 2.8% = 28 samples

Both Type X and Y defect = 0.3% = 3 samples

Both Type X and Z defect = 0.4% = 4 samples

Both Type Y and Z defect = 0.6% = 6 samples

Type X and Y and Z defect = 0.1%  = 1 sample

Venn Diagram is attached in the attachment below. Please refer to attachment for the Venn Diagram.

a) % age of samples containing none of these defects.

Solution:

Number of samples containing none of these defects = Total - Samples with defects

Number of samples containing none of these defects = 1000 - { (Type X) + (Type Y) + (Type Z) - (Both X and Y) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }

 Number of samples containing none of these defects = 1000 - { (21) + (24) +(28) - (3) -(4) - (6) + (1) }

Number of samples containing none of these defects = 1000 - 61

Number of samples containing none of these defects = 939

% age of samples containing none of these defects = 939/1000 x 100

% age of samples containing none of these defects = 93.9%

b) % age of samples containing at least one of these defects:

We have already calculated this above, number of samples containing at least on of these defects:

number of samples containing at least on of these defects = { (Type X) + (Type Y) + (Type Z) - (Both X and Y) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }

number of samples containing at least on of these defects = { (21) + (24) +(28) - (3) -(4) - (6) + (1) }  

number of samples containing at least on of these defects = 61

% age of samples containing at least one of these defects = 61/1000 x 100

% age of samples containing at least one of these defects = 6.1%

c) % age of samples free of type X and/or type Y defects.

For this find, we need to find the samples with only Z Type defect.

Number of Samples with Only Z type defects = { (Type Z) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }

Number of Samples with Only Z type defects = { (28)  -(4) - (6) + (1) }

Number of Samples with Only Z type defects = 19

Now, we also know the number of samples without any defects = 939

Now,

The number of samples free of type X and/or type Y defect = Sum of Number of Samples with Only Z type defects and number of samples without any defects

The number of samples free of type X and/or type Y defect = 19+939

The number of samples free of type X and/or type Y defect = 958

% age of samples free of type X and/or type Y defects = 958/1000 x 100

% age of samples free of type X and/or type Y defects = 95.8%

d) %age of samples with not more than 1 defect:

For this find, we need to find number of samples with only X type and with only type Y and with only type Z.

We have already found the number of samples with only Z type defect = 19

Now,

number of samples with only X type defect = { (Type X) - (Both X and Z) - (Both X and Y) + (All defects X and Y and Z) }

number of samples with only X type defect =  { (21)  -(4) - (3) + (1) }

number of samples with only X type defect =  15

Similarly,

number of samples with only Y type defect =  { (Type Y) - (Both Y and Z) - (Both X and Y) + (All defects X and Y and Z) }

number of samples with only Y type defect =  { (24)  -(6) - (3) + (1) }

number of samples with only Y type defect =  16

For,

samples with not more than 1 defect = number of samples with only Y type defect + number of samples with only X type defect + number of samples with only z type defect + number of samples without any defects

samples with not more than 1 defect = 939 + 16 + 15 + 19

samples with not more than 1 defect = 989

%age of samples with not more than 1 defect = 989/1000 x 100

%age of samples with not more than 1 defect = 98.9%

Answer:

Please check the attached graph.

Step-by-step explanation:

The slope-intercept form of the line equation

[tex]y = mx+b[/tex]

where

Given the system of equations

2x - 5y = -5    --- [Equation 1]

y = -3x + 1​      --- [Equation 2]

converting the equation 1 in the slope-intercept form

2x - 5y = -5

5y = 2x+5

divide both sides by 5

y = 2/5x + 1

Thus, the y-intercept  = 1

Now, please check the attached graph below.

The red line represents the equation  2x - 5y = -5

From the graph, it is clear that the red line has the y-intercept y = 1.

Now, considerng the equation 2

y = -3x + 1​  

comparing with the slope-intercept form of line equation

The y-intercept b = 1

Hence, the y-intercept of the line y = -3x + 1​ is y = 1

The green line represents the equation  y = -3x + 1​  

From the graph, it is clear that the green line has the y-intercept y = 1.

Point of Intersection:

It is clear from the graph:

As both lines meet or intersect at the point (0, 1).

Therefore, the point of intersection of both the lines is:

(x, y) = (0, 1)

As we know that the point of intersection of both the lines represents the solution to the system of equations.

Therefore, (0, 1) on the graph represents the solution to the system of equations.

The graph is attached below.