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3 years ago

A quarter, a nickel, and a dime are all flipped. how many possible outcomes are there?

Mathematics

1 answer:

AlekseyPX3 years ago

6 0

Possible outcomes

1. H H H

2. T T T

3. H T T

4. H H T

5. T T H

6. T H T

7. H T H

8. T H H

There are 8 possible outcomes.

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7. If F(x) = f(g(x)), where f(-2) = 8, f'(-2) = 4, f'(5) = 3, g(5)=-2, gʻ(5) = 6.. Find F'(5) <br>

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Answer:

I'm going to lay this out in a chart so it's a little easier to see:

F(x) = f(g(x))

x | f (x) | f ' (x) | g (x) | g ' (x)

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

-2 | 8 | 4 |

5 | | 3 | -2 | 6

Remember the chain rule, which says

(f (g (x))) ' = g ' (x) f ' (g (x))

When they ask for F ' (5), they are asking for (f (g (x))) ' when x = 5.

Using the chain rule, that's

F ' (5) = g ' (5) f ' (g (5))

We can simplify using the numbers provided.

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F ' (5) = 24

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

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Answer:

$\= x = 18.5$ , $\sigma = 5.15$

$15.505 < \mu < 21.495$

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

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