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Lina20 [59]
3 years ago
15

Solve for y 5x + y = -10

Mathematics
2 answers:
EleoNora [17]3 years ago
6 0
You would subtract 5x from y so y=5x-10
anygoal [31]3 years ago
6 0

Answer:

y=-(x/5)-2

Step-by-step explanation:

5x + y = -10

y=-(x/5)-2

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The amount of money Deon earns working at a store is given in the table.
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Answer:

C.

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The diagram shows part of a number line.Two of the fractions are not complete. Write the missing numerator in each box
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The first numerator is 3 and the second is 17 not 2
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PLEASE HELP. I don't know what to do. I tryed all the equations and the problem but none matched up. plz help
xz_007 [3.2K]

Answer:

178

1900 = 5(x+202)

Step-by-step explanation:

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1900 = 5 (202 + X)

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3 years ago
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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