All categories

2 years ago

84 times 0.24 area model

Mathematics

2 answers:

Paraphin [41]2 years ago

7 0

Can you explain more please i am willing to help but your question doesnt really have anything to do.

CaHeK987 [17]2 years ago

4 0

I'm not sure if I get what you mean, because I don't think I remeber what an area model is. But if it literally means making a model to describe the area, you could draw a square and put 84 on the length or the width and the o.24 on the other. Show that length times width = area. And then multiply. But if I am right, that's not what you are looking for, huh?

You might be interested in

It takes 62 pounds of seed to completely plant an 8-acre field. How many pounds of seed are needed per acre

Mandarinka [93]

7.75 pounds of seed are needed per acre

4 0

2 years ago

50 POINTS!! HELP!

Minchanka [31]

i dont have time to do all of it (sorry)

but this is how you do the point and slope questions:

use the equation: y-y1 = m(x - x1)

y1 is the y point you are given

x1 is the x point you are given

m is your slope

substitute all of those in from the question

the first one is:

y --4 = -5/6(x-8)

-- turns into a plus so the answer would be

y+4 = -5/6(x-8)

6 0

2 years ago

I need help please.

forsale [732]

Its letter D
7(19)-21 (x=19)
133-21=112 (RZ)
112+112=224 (RT)

4 0

2 years ago

Read 2 more answers

Example 1:

yawa3891 [41]

Answer:

Step-by-step explanation:

4 0

2 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

2 years ago