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3 years ago

15

Mattie uses the discriminant to determine the number of zeros the quadratic equation 0 = 3x2 – 7x + 4 has. Which bestdescribes t

he discriminant and the number of zeros?

1 answer:

Katyanochek1 [597]3 years ago

8 0

Discriminant Formula: b² - 4ac, with a = x^2 coefficient, b = x coefficient, and c = constant

So firstly, using our equation plug in the values into the discriminant formula and solve as such:

(-7)² - 4 × 3 × 4

49 - 48

1

So our discriminant is 1. <u>Since 1 is positive and a perfect square, this means that there are 2 real, rational solutions.</u>

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A coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that

mafiozo [28]

Answer:

(a) The significance level of the test is 0.002.

(b) The power of the test is 0.3487.

Step-by-step explanation:

We are given that a coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that the probability is not 0.5.

The test rejects the null hypothesis if either 0 or 10 heads are observed.

Let p = <u><em>probability of obtaining head.</em></u>

So, Null Hypothesis, $H_0$ : p = 0.5

Alternate Hypothesis, $H_A$ : p $\neq$ 0.5

(a) The significance level of the test which is represented by $\alpha$ is the probability of Type I error.

Type I error states the probability of rejecting the null hypothesis given the fact that the null hypothesis is true.

Here, the probability of rejecting the null hypothesis means we obtain the probability of observing either 0 or 10 heads, that is;

P(Type I error) = $\alpha$

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

Also, the event of obtaining heads when a coin is thrown 10 times can be considered as a binomial experiment.

So, X ~ Binom(n = 10, p = 0.5)

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

$\binom{10}{0}\times 0.5^{0} \times (1-0.5)^{10-0} +\binom{10}{10}\times 0.5^{10} \times (1-0.5)^{10-10}$ = $\alpha$

$(1\times 1\times 0.5^{10}) +(1 \times 0.5^{10} \times 0.5^{0})$ = $\alpha$

$\alpha$ = 0.0019

So, the significance level of the test is 0.002.

(b) It is stated that the probability of heads is 0.1, and we have to find the power of the test.

Here the Type II error is used which states the probability of accepting the null hypothesis given the fact that the null hypothesis is false.

Also, the power of the test is represented by (1 - $\beta$ ).

So, here, X ~ Binom(n = 10, p = 0.1)

$1-\beta$ = P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true)

$1-\beta$ = $\binom{10}{0}\times 0.1^{0} \times (1-0.1)^{10-0} +\binom{10}{10}\times 0.1^{10} \times (1-0.1)^{10-10}$

$1-\beta$ = $(1\times 1\times 0.9^{10}) +(1 \times 0.1^{10} \times 0.9^{0})$

$1-\beta$ = 0.3487

Hence, the power of the test is 0.3487.

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3 years ago

Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$

We get the $a^6b^2$ term when $k=2$ :

$k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2$

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1 year ago

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Answer:

I think the answer is 64 sqr foot try that and let me now

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3 years ago