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3 years ago

8

Compute the mean and variance of the following probability distribution. (Round your answers to 2 decimal places.) x P(x) 7 0.10

11 0.35 15 0.20 19 0.35

1 answer:

Alex777 [14]3 years ago

5 0

Answer:

14.2, 16.96

Step-by-step explanation:

given is a probability distribution of X

X is discrete taking values as 7, 11, 15 and 19

X 7 11 15 19 total

p 0.10 0.35 0.20 0.35 1

x*p 0.7 3.85 3.0 6.65 14.20

x^2*p 4.9 42.35 45 126.35 218 .60

Hence mean = sum of prob (x) = 14.20

$E(x^2) =218.60$

$Var(x) = E(x^2)-{E(x)}^2\\= 218.60-14.20^2\\= 16.96$

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Which theorem or postulate justifies this statement? angle6 cong angle13

dezoksy [38]

Answer:

C. Alternate Exterior Angles Theorem

Step-by-step explanation:

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2 years ago

A man weighing 800 N is hanging from a 2 mm diameter titanium wire with a cross-section of 3 x 10^-6 m^2. Will the wire break?

DENIUS [597]

Answer:

Since stress is greater than ultimate strength, the wire will break.

Step-by-step explanation:

The titanium wire is experimenting an axial load. Ultimate strength equals $2.20\times 10^{8}\,Pa$ . The wire shall break if and only if stress is at least equal to ultimate strength. The equation for axial stress ( $\sigma$ ), measured in pascals, in the wire with circular cross-section is:

$\sigma = \frac{4\cdot F}{\pi\cdot D^{2}}$ (1)

Where:

$F$ - Axial force, measured in newtons.

$D$ - Cross-section diameter, measured in meters.

Please notice that axial force is the weight of the man hanging from wire.

If we know that $F = 800\,N$ and $D = 0.002\,m$ , then the axial stress experimented by the titanium wire is:

$\sigma = \frac{4\cdot (800\,N)}{\pi\cdot (0.002\,m)^{2}}$

$\sigma \approx 2.546\times 10^{8}\,Pa$

Since stress is greater than ultimate strength, the wire will break.

3 0

2 years ago

What is the range of the function y = e Superscript 4 x

Alborosie

Answer:

A. Less than zero

Step-by-step explanation:

So If you graph the function, you'll see that it never actually crosses the y axis into the third or fourth quadrant, the quadrants where all y values become negative.

The answer A is also correct on Edge

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3 0

2 years ago

Read 2 more answers

Write a real-world problem that is solved by an equation and then solve (show work).

ololo11 [35]

A company is selling books. It has to pay $500 to start printing the books, and once they have done that, the books sell at $14.99 each. How many books must they sell to make a profit?

First we would model an equation. X will be the amount of books sold, and Y will be profits (in dollars obv). They had to pay $500 before they could start selling, so we must account for that too.

This equation would be $y=14.99x-500$ because for every book sold, X increases by 1, increasing Y by 14.99

The answer would be 34 books sold in order to turn a profit. (500/14.99=

6 0

2 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago