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3 years ago

5

The probability that a bakery has demand for 2 3 4 or 5 birthday cakes on any given day are 0.32, 0.24, 0.25 and 0.19 respective

ly. construct a probability distribution for this data

1 answer:

katrin2010 [14]3 years ago

4 0

This would just look like a bar graph.

On the horizontal axis, put your 4 different categories of 2, 3, 4, or 5.

On the vertical axis, label it by the percents: 0.10, 0.20, 0.30, 0.40

Then, make a bar at the correct height for each category.

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given the function f(x) = 2^x, find the value of f−1(32). (1 point) f−1(32) = 0 f−1(32) = 1 f−1(32) = 5 f−1(32) = 16

EastWind [94]

Answer:

5

Step-by-step explanation:

The inverse of $f(x)=2^x$ is $g(x)=\log_2(x)$ .

$g(32)=\log_2(32)$

$\log_2(32)=5$ since $2^5=32$ .

$g(32)=\log_2(32)=5$

Note: In general, the inverse of $f(x)=a^x$ is $g(x)=\log_a(x)$ .

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3 years ago

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olga_2 [115]

Answer:

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Set up and evaluate the optimization problems. (Enter your answers as comma-separated lists.) Find two positive integers such th

alexira [117]

Answer and Step-by-step explanation:

Let x and y be two positive integers and their sum is 14:

X + y = 14

And the sum of square of this number is:

f = x2 + y2

= x2+ (14 – x)2

Differentiate with respect to x, we get:

F’(x) = [ x2 + (14 – x)2]’ = 0

2x + 2(14-x)(-1) = 0

2x +( 28 – 2x)(-1) = 0

2x – 28 +2x = 0

2x + 2x = 28

4x = 28

X = 7

Hence, y = 14 – x = 14 -7 = 7

Now taking second derivative test:

F”(x) > 0

For x = y = 7,f reaches its maximum value:

(7)2 + (7)2 = 49 + 49

= 98

F at endpoints x Є [ 0, 14]

F(0) = 02 + (14 – 0)2

= 196

F(14) = (14)2 + (14 – 14)2

= 196

Hence the sum of squares of these numbers is minimum when x = y = 7

And maximum when numbers are 0 and 14.

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2 years ago

Which point would be a solution to the system of linear inequalities shown below? y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1 y&g

Sauron [17]

Answer:

$x < \frac{20}{17}$ and $y \ge \frac{-5}{17}$

Step-by-step explanation:

Given

$y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1$

Required

Find x and y

In the second equation. Assume that:

$y = \frac{3}{5}x - 1\\$

Substitute $y = \frac{3}{5}x - 1$ in the first equation

$y > 4x - 5$

$\frac{3}{5}x - 1 > 4x - 5$

Collect like terms

$\frac{3}{5}x - 4x > - 5 + 1$

$\frac{3}{5}x - 4x > -4$

Multiply through by 5

$3x - 20x > -20$

$-17x > -20$

Solve for x

$x < \frac{20}{17}$

Substitute this value of x in $y \ge \frac{3}{5}x - 1$

$y \ge \frac{3}{5}*\frac{20}{17} - 1$

$y \ge \frac{3}{1}*\frac{4}{17} - 1$

$y \ge \frac{12}{17} - 1$

$y \ge \frac{12- 17}{17}$

$y \ge \frac{-5}{17}$

3 0

3 years ago