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i think they meant this :)

 Construct a spinner with red, yellow, blue, and
green sectors, so the following probabilities are true.

• The probability of landing on red is 1/5
• The probability of landing on yellow is 50%.
• The probability of landing on blue is 1:10.
• The probability of landing on green is 2/10

Explain how you drew your spinner.

Your question reads

Construct a spinner with red, yellow, blue, and
green sectors, so the following probabilities are true.
• The probability of landing on red is .
• The probability of landing on yellow is 50%.
• The probability of landing on blue is 1:10.
• The probability of landing on green is .
Explain how you drew your spinner.

If you read closely it is empty after "red is" and "green is"

I'd like you to provide these values in order for me to help you.

Thanks.

i srsly dont understand :/

It's obvious you do since if you are using plain text it wouldn't be empty behind green and red.

i dont have a latex

please arrange your latex properly as we cannot see the prob. for green and red.

Ahhhh i see! 
thanks alan

Marvin roll a die. If he obtain a 5, he chooses a banana from box 1 where 4 are green and 6 are yellow. If he obtain a number that is not a 5, he chooses a banana from box 2 where 8 bananas are green and 3 bananas are yellow. What is the probability of obtaining a colored green banana?

P(green banana) = (1/6)*(4/10) + (5/6)*(8/11) 

I'll leave you to simplify the numbers.

collect all of the like terms:                        \(8x - 4 = 52\)

add 4 to both sides:                                          \(8x = 56\)

divide both sides by 8 to give x:                          \(x = 7\)

I interpreted the question as requiring a different experiment for each event.  However, if you need a single experiment, then how about something like:

Experiment:   

Throw a standard die.

Events:

a)  The probability of getting a number between 1 and 6 inclusive = 100%

b) The probability of getting an even number = 1/2

c) The probability of getting a 1 = 1/6

d) The probability of getting a 7 = 0

If there is some other interpretation of your question you will need to clarify!

Either increasing the radius or the height of a cylinder by six inches will result in the same volume. The original height of the cylinder is two inches. What is the original radius?  

Volume of a Cylinder  =  π • r² • h   

If you increase the radius by 6 the new volume will be   π • (r+6)² • 2  

If you increase the height by 6 the new volume will be   π • r² • (2+6)  

Set the new volumes equal and solve for r  

                                                                                     π • (r² + 12r + 36) • 2  =  π • r² • 8   

First, let's divide both sides by π                                        (r² + 12r + 36) • 2  =        r² • 8   

Next, let's divide both sides by 2                                          r² + 12r + 36         =        r² • 4   

Subtract r² from both sides                                                         12r + 36         =        3r²     

Subtract (12r + 36) from both sides                                                    0             =        3r² – 12r – 36

Switch sides, just cuz I prefer the zero on the right             3r² – 12r – 36  =  0   

Factor the quadratic                                                            (3r + 6)(r – 6)  =  0    This is the hardest step, IMHO.

Set each factor equal to zero                                                3r + 6   =  0 

                                                                                                      r  =  –2             Discard... can't have negative radius.

                                                                                                 r – 6  =  0 

                                                                                                       r  =  6              The original radius was 6.   . 

_.

The probability of throwing a 1 with a standard six-sided die is 1/6 (which I assume is what part  c) is asking for).

For part a) the probability of throwing a 1 with a six-sided die with 1 on every face is 100%.

etc.

I don't understand make it clear

because for all you did 

"Probability of throwing a 1."

and I don't see part c

Let x and y be integers. Show that 9x+5y is divisible by 19 if and only if x+9y is divisible by 19.

As follows:

Given the factors shown in the previous answer, 7 has no common factors other than itself, therefore you will have to locate the 7 between the only other two factors that have 7 as a factor (i.e., are divisible by 7). 

_.

In the figure triangle ABC is a right triangle with legs AB = 6 and AC = 8.

A square is drawn as shown, with a side along BC and corners on AB and AC.

Find the length of the side of the square.

\(\begin{array}{|rclrclrcl|} \hline \tan(C) &=& \dfrac{x}{p} & && & \tan(C) &=& \dfrac{6}{8} \\ &&& \dfrac{x}{p} &=& \dfrac{6}{8}& \\ && &\dfrac{p}{x} &=& \dfrac{8}{6}& \\ && &p &=& \dfrac{8x}{6}& \\ && &\mathbf{p} &=& \mathbf{\dfrac{4x}{3}}& \\ \hline \end{array} \begin{array}{|rclrclrcl|} \hline \tan(B) &=& \dfrac{x}{q} & && & \tan(B) &=& \dfrac{8}{6} \\ &&& \dfrac{x}{q} &=& \dfrac{8}{6}& \\ && &\dfrac{q}{x} &=& \dfrac{6}{8}& \\ && &p &=& \dfrac{6x}{8}& \\ && &\mathbf{q} &=& \mathbf{\dfrac{3x}{4}}& \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline p+x+q &=& 10 \\\\ \dfrac{3x}{4}+x+\dfrac{4x}{3} &=& 10 \\\\ x*\left( \dfrac{3}{4}+1+\dfrac{4}{3} \right) &=& 10 \\\\ x* \dfrac{37}{12} &=& 10 \\\\ x &=& 10*\dfrac{12}{37} \\\\ x &=& \dfrac{120}{37} \\\\ \mathbf{x} &=& \mathbf{3.\overline{243}} \\ \hline \end{array}\)

Let the precise time of arrival =T

45[T + 5/60] =63[T - 5/60], solve for T

T = ½ an hour

Sub the T into the above equation to get one-way distance =26.25 miles

26.25 x 2 =52.50 miles - distance back and forth.

So, to arrive at precise time at the office, he must drive at: 52.5 miles/ 1 hour=52.5 mph - his average speed.

Awesome, Thankyou so much!

Did you do it with the square?

As follows:

Thank you so much!!!

Side AC = 8

Let  x be a side of the square

Triangles  JCG and AJI are similar.

The sine of an angle JCG = 0.8

The cosine of an angle AJI = 0.6

0.8 * x + x / 0.6 = 8                                                                                                                                                                                                                               

0.8x + 1.666667x = 8

2.466667x = 8   

Divide both sides by 2.466667

2.466667x / 2.466667 = 8 / 2.466667

x = 3.243243   

P(2≤X≤5) is given by the area under the curve between 2 and 5.  This is:

(1/2)*(4-2)*(0.2-0.05) + (1/2)*(5-4)*(0.15-0.05) + (5-2)*0.05  =  0.35