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3 years ago

Solve the proportion. Show your work. PLEASE

Mathematics

2 answers:

dolphi86 [110]3 years ago

8 0

Answer:

Hey, i know its been like 2 years.

Step-by-step explanation:

But is there anyway you can send me the final test for ashworth geometry part 2?

Dmitrij [34]3 years ago

5 0

$\bf \cfrac{3}{x+1}=\cfrac{5}{11}\implies (11)3=(x+1)5\implies 33=5x+5
\\\\\\
28=5x\implies \cfrac{28}{5}=x\implies 5\frac{3}{5}=x$

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I running track is 400 meters. if i make 4 laps, what is the distance that i traveled?

IgorLugansk [536]

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3 years ago

Rectangle has a vertices at (-1,4), (-1,-1) and (4, -1) where is the fourth vertex located

erica [24]

Answer:

(4,4)

Step-by-step explanation:

1. I drew a grid and plotted the 3 points I already knew

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3 years ago

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romanna [79]

Answer:

Step-by-step explanation

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6 0

3 years ago

Read 2 more answers

X^2 + y^2 = 8<br> X-y=0<br> Select all of the following that are solutions to the system shown

ludmilkaskok [199]

X^2 + y^2 = 8
X-y=0 so x = y

replace x = y into X^2 + y^2 = 8

y^2 + y^2 = 8

2y^2 = 8

y^2 = 8/2

y^2 = 4

y = - 2 and y = 2

because x = y

so x = - 2 and x = 2

solutions:

x= - 2 and x = + 2

y= - 2 and y = + 2

5 0

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

2 years ago