A probability experiment consists of rolling a fair 10 sided die. probability of rolling a 1

(10pts) A train left a station at 8:00 am

dolphi86 [110]

Answer:

12:00 i think

Step-by-step explanation:

hope this helps

6 0

3 years ago

Please help me .....................

Flauer [41]

Nansbdjs and uiaishdwueu828

6 0

3 years ago

Evaluate the sum of the following finite geometric series. 5+15+45 +135+ . . . + 1215 Sn=1

Nata [24]

Answer:

1820

Step-by-step explanation:

The sum of the first n terms of a geometric series with common ratio r and first term a is:

a(1-r^n)/(1-r). Plugging this in 5(1-3^6)/(1-3)=1820

5 0

3 years ago

Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall f

spin [16.1K]

Answer:

Part 1) 0.50 inches

Part 2) 1.14 inches

Part 3)

a) 1.25 inches

b) -0.32 inches

Step-by-step explanation:

The complete question in the attached figure

Part 1) Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall for the last June

Let

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

The difference between the average rainfall and the actual rainfall for the last June is given by the expression

$(x-y)$

we have

$x=0.67\ in\\y=0.17\ in$

substitute

$0.67-0.17=0.50\ in$

Part 2) The departure from the average rainfall last July was -0.36 inches. How much rain fell last July?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

so

$D=y-x$

we have

$D=-0.36\ in\\x=1.5\ in$

substitute the given values

$-0.36=y-1.5$

solve for y

$y=1.5-0.36\\y=1.14\ in$

Part 3)

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Part b) What would the departure from the average be in August in that

situation?

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Find the average rainfall for these three months

Adds the average rainfall and divide by 3

$(0.67+1.5+1.57)/3=1.25\ in$

Part b) What would the departure from the average be in August in that

situation?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

so

$D=y-x$

we have

$x=1.57\ in\\y=1.25\ in$

substitute the given values

$D=1.25-1.57=-0.32\ in$

7 0

4 years ago

The x-y is the answer can someone please explain to me how to do

guapka [62]

Answer:

Step-by-step explanation:

$\frac{\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}-\frac{\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}}\\\\=\frac{\sqrt{x-y}*(\sqrt{x+y}+\sqrt{x-y})-(\sqrt{x-y})*(\sqrt{x+y}-\sqrt{x-y})}{(\sqrt{x+y}-\sqrt{x-y})*(\sqrt{x+y}+\sqrt{x-y})}\\\\=\frac{\sqrt{x-y}*(\sqrt{x+y}+\sqrt{x-y}-\sqrt{x+y}+\sqrt{x-y})}{(\sqrt{x+y})^2-(\sqrt{x-y)^{2}}}\\\\=\frac{\sqrt{x-y}*2\sqrt{x-y}}{x+y-x+y}\\\\=2*\frac{x-y}{2y}=\frac{x-y}{y}\\\\(\frac{\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}-\frac{\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}})*y=\frac{x-y}{y}*y\\\\=x-y$

4 0

4 years ago

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