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2 years ago

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An arithmetic series contains 20 numbers. The first number is 102. The last number is 159. Which expression represents the sum o

f the series?

2 answers:

Lina20 [59]2 years ago

5 0

Answer: The expression is,

$\frac{20}{2}(102+159)$

Step-by-step explanation:

Since, the sum of an A.P. is,

$S_n=\frac{n}{2}(a+a_n)$

Where, n is the number of terms,

a is the first term of the A.P.,

$a_n$ is the last term of the A.P.,

Here, we have given,

a = 102, $a_n$ = 159, and n = 20,

Thus, the sum of the given A.P. is,

$S_20=\frac{20}{2}(102+159)$

Which is the required expression.

kkurt [141]2 years ago

3 0

If $a_1=102$ is the first term in the series, then the next term is $a_2=a_1+k$ for some fixed $k$ , the term after that is $a_3=a_2+k=a_1+2k$ , and so on. So the 20th term in the series would be $a_{20}=a_1+(20-1)k=102+19k=159$ .

We have enough info to find $k$ :

$102+19k=159\implies19k=57\implies k=3$

So the sum of the series is given by

$\displaystyle\sum_{n=1}^{20}(a_1+3(n-1))=\sum_{n=1}^{20}(3n+99)=2610$

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A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o

NeX [460]

Answer:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ 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 adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

$\hat p=\frac{115}{180}=0.639$ estimated proportion of adults with the medication was effective

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is less than 0.65.:

Null hypothesis: $p \geq 0.65$

Alternative hypothesis: $p < 0.65$

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

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

Calculate the statistic

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

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

Statistical decision

It's important to refresh the p value method or p value approach . "This method 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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ 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 adults with the medication was effective is not significantly less than 0.65

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