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mylen [45]
3 years ago
11

What is the percent of decrease between 80 and 64

Mathematics
1 answer:
8090 [49]3 years ago
6 0
80 - 64 = 16
16 ÷ 80= 0.2
0.2= 20%
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8y-10=6y-2 help me please I don't get this
murzikaleks [220]
8y-10=6y-2 

8y-10=6y-2:y  | Solve \ for \ y| 

8y=6y-2+10
 
  |Add \ 10 \ to \ both \ sides| 


8y=6y+8
 
 


8y-6y=8
 
  |Subtract \ 6y \ from \ both \ sides|

2y=8 

y=8/2   |Divide \ both  \ sides \ by \ 2|

y=4 

\star \ Hence \ solved

6 0
3 years ago
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Which rigid transformation would map ΔAQR to ΔAKP? a rotation about point A a reflection across the line containing AR a reflect
Jobisdone [24]

Answer:

a rotation about point A

Step-by-step explanation:

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3 years ago
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2.- Supóngase que los diámetros de los tornillos fabricados por una compañía están distribuidos normalmente con una media de 0.2
Sergio [31]

Answer:

7.30167%

Step-by-step explanation:

Usando la fórmula de puntuación z

z = (x-μ) / σ, donde x es la puntuación bruta, μ es la media de la población y σ es la desviación estándar de la población

Para x <0.20 pulgadas

z = 0.20 - 0.25 / 0.02

z = -2.5

Valor de probabilidad de Z-Table:

P (x <0.20) = 0.0062097

Para x> 0.28 pulgadas

z = 0.28 - 0.20 / 0.02

z = 1.5

Valor de probabilidad de Z-Table:

P (x <0.28) = 0.93319

P (x> 0.28) = 1 - P (x <0.28) = 0.066807

La probabilidad de que se produzcan tornillos defectuosos cuando el tornillo se considera defectuoso si su diámetro es inferior a 0.20 pulgadas o superior a 0.28 pulgadas es

P (x <0.20) + P (x> 0.28)

= 0.0062097 + 0.066807

= 0.0730167

Conversión a porcentaje

= 0.0730167 × 100

= 7.30167%

El porcentaje de tornillos defectuosos producidos es

7.30167%

3 0
2 years ago
An ice cream sandwich has a length of 4 inches, width of 2 1/2 inches, and height of 3/4 inches.What is the volume of the ice cr
dsp73

Answer:

7.2 is your answer:)

Step-by-step explanation:

8 0
3 years ago
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In a study of the accuracy of fast food​ drive-through orders, one restaurant had orders that were not accurate among orders obs
Evgesh-ka [11]

Answer:Null hypothesis:  

Alternative hypothesis:  

 

 

The p value obtained was a very high value and using the significance level given  we have  so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .  

and the accuracy of the test yes is acceptable since the p value obtained is large enough to fail to reject the null hypothesis

Step-by-step explanation:

Data given and notation n  

n=367 represent the random sample taken

X=32 represent the orders that were not accurate

estimated proportion of orders that were not accurate

is the value that we want to test

represent the significance level

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z would represent the statistic (variable of interest)

 

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion  is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

 

Statistical decision  

It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided . The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

 

So the p value obtained was a very high value and using the significance level given  we have  so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .  

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3 years ago
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