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Grace [21]
1 year ago
8

Find the sum 112.1+ 1.2

Mathematics
2 answers:
dimaraw [331]1 year ago
8 0
The answer is 113.3 hope this helps
ozzi1 year ago
7 0
112.1 plus 1.2 is 113.3
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Radium-226, an isotope of radium, has a half-life of 1,601 years. Suppose your chemistry teacher has a 50-gram sample of radium-
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We know that
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An=A0*(0.5)<span>^[t/h)]
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h------------> is the half-life of the decaying quantity

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h=1601 years
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the answer is 47.88 g
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There is a harvest festival each year on the planet Bozone. During that time, most of the Bozone residents sit down to rejoice a
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The price would be decreased by 18 bozats

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A member of a student team playing an interactive marketing game received the fol- lowing computer output when studying the rela
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Answer:

p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91

So on this case for the significance level assumed \alpha=0.05 we see that p_v >\alpha so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

Step-by-step explanation:

Let's suppose that we have the following linear model:

y= \beta_o +\beta_1 X

Where Y is the dependent variable and X the independent variable. \beta_0 represent the intercept and \beta_1 the slope.  

In order to estimate the coefficients \beta_0 ,\beta_1 we can use least squares procedure.  

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: \beta_1 = 0

Alternative hypothesis: \beta_1 \neq 0

Or in other words we want to check is our slope is significant (X have an effect in the Y variable )

In order to conduct this test we are assuming the following conditions:

a) We have linear relationship between Y and X

b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable

c) We assume that the Y values are independent and the distribution of Y is normal  

The significance level assumed on this case is \alpha=0.05

The standard error for the slope is given by this formula:

SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}

Th degrees of freedom for a linear regression is given by df=n-2 since we need to estimate the value for the slope and the intercept.  

In order to test the hypothesis the statistic is given by:

t=\frac{\hat \beta_1}{SE_{\beta_1}}

The p value on this case would be given by:

p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91

So on this case for the significance level assumed \alpha=0.05 we see that p_v >\alpha so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

3 0
2 years ago
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