Map of a mall is overlaid with a numeric grid. The kiosk for the

Suppose the time to process a loan application follows a uniform distribution over the range 77 to 1515 days. What is the probab

Delicious77 [7]

Answer:

The probability is approximately 0.2222

Step-by-step explanation:

The correct question is as follows;

Suppose the time to process a loan application follows a uniform distribution over the range 7 to 15 days. What is the probability that a randomly selected loan application takes longer than 14 days to process

Please check attachment for solution to question

5 0

3 years ago

Basketballs sell for 27.99 each at the super D and for 21.59 each at the bargain spot. If the coach buys a dozen basketballs, ho

Mazyrski [523]

What I did was 27.99-21.59 which equals 6.40 and then times that by 12 which brings you to 76.80

3 0

3 years ago

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Find the formula for the area of the shaded region in the figure.<br> x² = 4py<br> ہے

Alexandra [31]

The formula for the area of the shaded region in the figure.

x² = 4py is mathematically given as

$A=\frac{8}{3} \sqrt{p h} \cdot h$

<h3>What is the formula for the area of the shaded region in the figure?</h3>

Parameters

$x^{2}=4 p y$

$y=h$

$\therefore x^{2}=4 p h \Rightarrow x=2 \sqrt{p h} \\$

Generally, the equation for the Area is mathematically given as

$& A=2 \int_{0}^{2 \sqrt{p h}}\left(h-\frac{x^{2}}{4 p}\right) d x \\\\& A=2\left(h x-\frac{x^{3}}{12 p}\right]_{0}^{2 \sqrt{p h}} \\\\& A=2\left[\left[h 2 \sqrt{p h}-\frac{1}{12 p} \cdot(2 \sqrt{p h})^{3}-0\right]\right. \\$

$&A=2\left[2 h^{3 / 2} \sqrt{p}-\frac{8(\sqrt{p})^{3}(\sqrt{h})^{3}}{12 p}-0\right] \\\\&A=2\left[2 h^{k / 2} \sqrt{p}-\frac{2}{3} \sqrt{p} \cdot h^{3 / 2}\right] \\\\&A=2 \times \frac{4 \sqrt{p} \cdot h^{3 / 2}}{3} \\\\&A=\frac{8}{3} \cdot \sqrt{p} \cdot \sqrt{h} \cdot h \\\\&A=\frac{8}{3} \sqrt{p h} \cdot h$