81 is divisible by? 1. 3 2. 4 3. 6 4. 8

Which of the following equations represents a line that is parallel to the line that passes through the points below?

Pepsi [2]

Answer:

See below

Step-by-step explanation:

Two lines are parallel if they have the same slope but different y-intercepts. So the slope that can be formed given the points (-3,6) and (9,2) is (9-(-3))/(2-6)=12/-4=-3

With y=-3x+b, we need b, which can be found by plugging in either point:

2=-3(9)+b

2=-27+b

29=b

So the y-intercept therefore cannot be 29 but it can be any real number.

So the equation y=-3x+2 works, y=-3x+3, y=-3x+4, and so on....

3 0

3 years ago

What cannot be used as evidence in a proof

gtnhenbr [62]

K12? Me too.
It would be educated guess. I just finished that practice cause I'm catching up :) Good luck!

8 0

3 years ago

Find the measure of angle b if angle a is 22 degrees and the whole angle is 300 degrees

Dima020 [189]

Answer:

$a = 278^{\circ}$

Step-by-step explanation:

Given

$b = 22^{\circ}$

$Whole\ Angle = 300^{\circ}$

Required

Find the measure of a

Since there is no diagram to support the question, we'll assume that:

$a + b = Whole\ Angle$

This gives:

$a + 22^{\circ} = 300^{\circ}$

Subtract 22 from both sides

$a + 22^{\circ}-22^{\circ} = 300^{\circ}-22^{\circ}$

$a = 300^{\circ}-22^{\circ}$

$a = 278^{\circ}$

6 0

3 years ago

The maximum weight for a truck on the New York State Thruway is 40 tons. How many pounds is this?

galben [10]

40 tons is 80,000 pounds

5 0

4 years ago

Read 2 more answers

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

b)

- 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

b)

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