Write a fraction that is less than 5/6 but has a denomator of 8

The annual rainfall (in inches) in a certain region is normally distributed with = 40 and = 4. What is the probability that star

sdas [7]

Answer:

0.93970

Step-by-step explanation:

Solution:-

- Denote a random variable "X" The annual rainfall (in inches) in a certain region . The random variable follows a normal distribution with parameters mean ( μ ) and standard deviation ( σ ) as follows:

X ~ Norm ( μ , σ^2 )

X ~ Norm ( 40 , 4^2 ).

- The probability that it rains more than 50 inches in that certain region is defined by:

P ( X > 50 )

- We will standardize our test value and compute the Z-score:

P ( Z > ( x - μ ) / σ )

Where, x : The test value

P ( Z > ( 50 - 40 ) / 4 )

P ( Z > 2.5 )

- Then use the Z-standardize tables for the following probability:

P ( Z < 2.5 ) = 0.0062

Therefore, P ( X > 50 ) = 0.0062

- The probability that it rains in a certain region above 50 inches annually. is defined by:

q = 0.0062 ,

- The probability that it rains in a certain region rains below 50 inches annually. is defined by:

1 - q = 0.9938

n = 10 years ..... Sample of n years taken

- The random variable "Y" follows binomial distribution for the number of years t it takes to rain over 50 inches.

Y ~ Bin ( 0.9938 , 0.0062 )

- The probability that it takes t = 10 years for it to rain:

= 10C10* ( 0.9938 )^10 * ( 0.0062 )^0

= ( 0.9938 )^10

= 0.93970

3 0

3 years ago

If the geometric mean of a and 34 is 6 sqrt(17) find the value of a

Alborosie

Using the geometric mean concept, it is found that the value of a is 18.

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The geometric mean, of a data-set of n elements, $(n_1, n_2, ..., n_n)$ , is given by:

$G = \sqrt[n]{n_1 \times n_2 \times ... \times n_n}$

That is, the nth root of the multiplication of all elements.

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

In this question:

Two elements(n = 2), a and 34.
$G = 6\sqrt{17}$

Thus:

$\sqrt{34a} = 6\sqrt{17}$

We find the square of each side, so:

$(\sqrt{34a})^2 = (6\sqrt{17})^2$

$34a = 36\times17$

Simplifying both sides by 17:

$2a = 36$

$a = \frac{36}{2}$

$a = 18$

The value of a is 18.

A similar example is given at brainly.com/question/15010240

3 0

3 years ago

There are two pizzas. Conor ate 1⁄4 of a pizza, Brandon 2⁄8, Tyler 3⁄4, and Audrey4⁄8. Who ate the most of the two pizzas?

OLEGan [10]

B. Tyler ate the most of the two pizzas.

6 0

3 years ago

Read 2 more answers

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is: