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3 years ago

Solve : 5x - 7y = 12 5y - 2x = - 7

2 answers:

6 0

5x - 7y = 12
5y - 2x = -7
5x - 7y = 12
-2x + 5y = -7
3x + 2y = 5
3(2) + 2(-1) = 5
6 + (-1) = 5
6 - 1 = 5
5 = 5

Afina-wow [57]3 years ago

3 0

There is one solution:

use elimination method

5x - 7y = 12
5y - 2x = -7 ---> change equation around

5x - 7y = 12
-2x + 5y = -7

2 x ( 5x - 7y) = 2 x (12) multiply both sides by 2
5 x (-2x +5y) = 5 x (-7)

this give you
10x - 14y =24
-10x+25y = -35 add down
-----------------------------
11y = - 11 x is eliminated to find y value

y = -1 input to one of the original equations

5(-1) - 2x = -7

-5 - 2x = -7
+5 +5 add 5 to both sides
----------------------------
-2x = -2
x = 1

your coordinates for when they intersect is at (1, -1)
one solution

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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

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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 = probability of obtaining head.

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) = $\alpha$

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

$\alpha$ = 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.

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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)

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

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

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