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svetlana [45]
3 years ago
12

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th

e highest bid in excess of $9,500 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $9,500 and $14,700. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?
Mathematics
1 answer:
Karolina [17]3 years ago
5 0

Answer:

Probability (bid accepted) = 0.48

Step-by-step explanation:

Probability density is given byF(y)= 1/(b-a)

a=9500

b= 14700

F(y)= 1/(14700-9500) =1/5200=0.00019

Probability (bid accepted)= (12000-9500)÷1/5200

P( bid accepted) = 2500×0.00019=0.475 approximately 0.48

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vlabodo [156]

Answer:

-11

Step-by-step explanation:

-6+(-1)+(-4)=-11

I hope this helps, I also think this is the answer!.

7 0
3 years ago
Last question please
MatroZZZ [7]

Let's try an make a formula to make things easy.

_________________________________________________

Look at Andrew's pattern, it goes up 6 every time. Lets look at the 6 times table

6 times table:           6, 12, 18, 24, ...

____________________________________

First think, how would you work out the <u>7</u>th term in the <u>6</u>th times table?

You would do <u>6</u> times <u>7</u>.

So the formula for the 6th times table is:

6n     (where n is the term you want to find out)

_______________________________________

So to get the 9th term in the 6 times table, you would replace the n with 9, like so:

9th term in 6 times table = 6 x 9 = 54

______________________________________

Now lets compare the 6 times table to Andrew's pattern:

6 times table:           6, 12, 18, 24,

Andrew's pattern:     3, 9, 15, 21,

_____________________________________

If you look, you can see that Andrew's pattern is the 6 times table minus 3!

So what would be the formula for Andrew pattern?

---> it would be:  6n - 3   (since it's the 6 times table but minus 3)

_____________________________________

We can use this formula to easily work out the 20th term in Andrew's pattern!:

20th term = 6 x 20 - 3 = 120 - 3 = <u>117</u>    (remember, replace the n with 20)

______________________________________

Now let's make a formula for Ben's pattern!

Ben's pattern goes up in 5s, so lets compare it to the 5 times table:

5 times table:       5, 10, 15, 20

Ben's pattern:       3, 8, 13, 18,

The formula for the 5 times table is 5n, and Ben's pattern is the 5 times table but minus 2.

So the formula for Ben's pattern is: 5n - 2

__________________________________

<u>Now swap out the n for 20</u>  (to get the 20th term):

20th term = 5 x 20 - 2 = 100 - 2 = <u>98</u>

_________________________________________

So Andrew's 20th term is <u>117</u>, and Ben's 20th term is <u>98</u>.

All we do now is subtract them to find the difference:

117 - 98 = <u>19</u>

- - - - - - - - - - - - - - - - - - - - - - - - -

Answer:

Ben's 20th term is smaller by 19!

- - - - - - - - - - - - - - - - - - - - - - - - - -

5 0
3 years ago
Find the area of the circle.leave your answer in terms of
Yakvenalex [24]
Area is 3.6*3.6*pi
Area is 12.96 pi
5 0
3 years ago
Please help
Alekssandra [29.7K]

Answer:

25 in x 15 in

Step-by-step explanation:

Given:

  • Length = 3/5 the width
  • Area = 375 in²

Let width = x

Therefore, length = 3/5 x

First create an equation for the area of the picture based on the given information for its width and length:

\begin{aligned} \implies \textsf{Area of original picture} & = \sf width \times length\\& = x\left(\dfrac{3}{5}x\right)\\& = \dfrac{3}{5}x^2\end{aligned}

We are told the area of the enlarged picture is 375 in².  Therefore, substitute this into the equation and solve for x to find the width of the enlarged picture:

\begin{aligned}\textsf{Area} & = 375\\ \implies \dfrac{3}{5}x^2 & = 375\\ x^2 & =375 \cdot \dfrac{5}{3}\\ x^2 & =625\\ x & = \sqrt{625}\\ x& = 25\end{aligned}

Therefore, the width of the enlarged picture is 25 in.

Substitute the found value of x into the expression for length to find the length of the enlarged picture:

\begin{aligned}\sf Length & = \dfrac{3}{5}x\\\\\implies \sf Length & = \dfrac{3}{5}(25)\\\\& = 15\: \sf in\end{aligned}

Therefore, the dimensions of the enlarged picture are <u>25 in x 15 in</u>.  The width is 25 in and the length is 15 in, as the length is 3/5 of the width.

5 0
2 years ago
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