Prove that the roots of the equation X^2 - 2px + p^2 - q^2 =0 where p and q are constants, are real.

Mathematics

1 answer:

lora16 [44]3 years ago

3 0

Answer:

Step-by-step explanation:

Sorry I am not sure and don’t want to give you the wrong answer sorry again

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Direct mail advertisers send solicitations ("junk mail") to thousands of potential customers in the hope that some will buy the

harina [27]

Answer:

The interval is (.0817, .118).

Step-by-step explanation:

Let's make this interval! The easy way is to go to STAT->TEST->A, 1-prop z-int, but I will show the long way.

1. Conditions

First we must check the conditions.

Randomization condition: The sample is given to be random, so we can assume independence.

Success/failure condition: Both np>10 and nq>10 must be met.

np = 1030(103/1030) = 103 ✓
nq = 1030(927/1030) = 927 ✓

10% condition: The sample cannot be greater than 10% of the population.

10n < N, 10(1030) < 200000 ✓

All conditions are met, so we can use a 1-proportion z-interval to represent the data.

2. Mechanics

Find $\hat{p}$ , the sample proportion.

103/1030 = .1

$\hat{q}$ , then, the probability of not p, is 1 - .1 = .9

Find z* (z star), the critical z-value.

You can do this on your calculator using 2nd->VARS->3 (or you can memorize it - it's 1.96. Winky face.) This is called the invNorm function, that takes the left side area under the normal curve. But what do we plug in to the invNorm function?

invNorm(.975) = 1.96

Plug values into the equation.

$\hat{p}$ ± z* $\sqrt{\frac{\hat{p}\hat{q}}{n} }$

.1 ± 1.96 $\sqrt{\frac{.1*.9}{1030} }$

.1 ± 1.96(.00935)

.1 - .0183 = .0817

.1 + .0183 = .118

(.0817, .118)

3. Conclusion

Based on this sample, we are 95% confident that the proportion of people the company contacts who may buy something is between .0817 and .118. That isn't very likely, but maybe they'll make a little profit!

5 0

3 years ago

In the expansion of

Mumz [18]

By the binomial theorem, the $x^2$ term in the expansion of $\left(1-\frac x4\right)^n$ is