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liq [111]
3 years ago
9

Research was done to study the facial expressions of people, since facial expression clearly shows the emotional state of a pers

on. Researchers found that people who smiled a lot lived longer than those who didn't smile much.
Which statement can be inferred from the passage?


Smiling prolongs human life.



Smiling is associated with human longevity.



Smiling decreases a person's life span.



There is no link between smiling and human longevity.
Mathematics
2 answers:
finlep [7]3 years ago
8 0
B 

<span>A is wrong because you can't infer a cause and effect relationship between smiling and long life (other factors might be at work) </span>

<span>C and D are clearly wrong

</span>
Andrej [43]3 years ago
8 0

<em>Mark Me Brainliest !</em>

Question :

Which statement can be inferred from the passage?

Answer:

A.  Smiling prolongs human life.

<em> B [ Smiling is associated with human longevity ]</em>

C.  Smiling decreases a person's life span.

D.  There is no link between smiling and human longevity.


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Step-by-step explanation:

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A bicycle manufacturing company makes a particular type of bike. Each child bike requires 4 hours to build and 4 hours to test.
Verdich [7]

 the first option is the answer

Step-by-step explanation:

4c + 6a <= 120 [building hours]

4c + 4a <= 100 [testing hours]

4 * 20 + 6 * 6 <= 120

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true, they have enough building hours

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If the legs of an isosceles right triangle have a length of 15 StartRoot 2 EndRoot ft, what is the length of the hypotenuse?
lorasvet [3.4K]

Answer:

30 ft

Step-by-step explanation:

a² + b² = c²

(15sqrt(2))² + (15sqrt(2))² = c²

225 * 2 + 225 * 2 = c²

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c = sqrt(900)

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A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar
Arisa [49]

Answer:

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

n=17 represent the sample selected

\alpha=0.1 represent the significance

s^2 =4.7^2 =22.09 represent the sample variance

\sigma^2_0 =3.5^2 =12.15 represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 12.15

Alternative hypothesis: \sigma^2 \neq 12.15

The statistic for this case is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

The degrees of freedom are:

df =n-1= 17-1=16

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0
3 years ago
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