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3 years ago

7th grade work (: haha

Mathematics

1 answer:

Ray Of Light [21]3 years ago

5 0

Answer:

Option B $-(-8\frac{1}{4})$

Step-by-step explanation:

we know that

The opposite or additive inverse of a number a is the number that, when added to a, yields zero.

The opposite of a is -a

In this problem we have

$8\frac{1}{4}$

therefore

The opposite is

$-8\frac{1}{4}$

and the opposite of the opposite is

$-(-8\frac{1}{4})$

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Julio says, "If you subtract 17 from my number and multiply the difference by negative 4, the result is negative 92." What is

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"If you subtract 17 from my number and multiply the difference by negative 4, the result is negative 92.

n= number

(n-17)*(-4) = -92

distribute

-4 *n - 17*(-4) = -92

-4n +68 = -92

subtract 68 from each side

-4n +68-68 = -92 -68

-4n = -160

divide each side by -4

-4n/-4 = -160/-4

n = 40

Julio's number is 40

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What is the solution to this system of equations?

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Find the greatest common factor of 14 and 28

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3 years ago

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3 years ago

A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o

ryzh [129]

Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑ $_{x=0$ b( x, n, p )

= ³∑ $_{x=0$ b( x, 10, 0.6 )

= ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.6)^x$ $( 1 - 0.6 )^{10-x$

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.3 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.3)^x$ $( 1 - 0.3 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.3)^x$ $( 1 - 0.3 )^{10-x$

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.4 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.4)^x$ $( 1 - 0.4 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.4)^x$ $( 1 - 0.4 )^{10-x$

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.5 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.5)^x$ $( 1 - 0.5 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.5)^x$ $( 1 - 0.5 )^{10-x$

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

3 0

3 years ago