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4 years ago

Which sequence is modeled by the graph below?

Mathematics

1 answer:

Maru [420]4 years ago

3 0

We have

a2 = 4

a3 = 2

a4 = 1

calculate the difference

a3 - a2 = -2

a4 - a3 = -1

The differences are not equal---------- >the sequence is not arithmetic.

Getting the ratio:

a3/a2 = 0.5

a4/a3 = 0.5

The common ratio is 0.5. Using the general form for a geometric series:

an = a1 r^(n-1)

for n = 2---------------- > a2=4

4 = a1 (0.5)^(2-1)

a1 = 8

therefore

an = 8 (1/2)^(n-1)

the answer is an = 8 (1/2)^(n-1)

the answer is none of those indicated in the question

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Simplify for a = 3, b= -4 and c=-1.<br> 16a - 3c + 5b =

-Dominant- [34]

Answer:

Step-by-step explanation:

16(3)-3(-1)+5(-4)

48+3-20

51-21

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16. The gravitational potential energy of an object is always measured

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Surface of earth i guess

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The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where paramet

never [62]

Answer:

a) $t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$

The degrees of freedom are given by:

$df=n-1=45-1=44$

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

b) $p_v =P(t_{(44)}>2.744)=0.0044$

Step-by-step explanation:

Information given

$\bar X=18.4$ represent the sample mean

$s=2.2$ represent the sample standard deviation

$n=45$ sample size

$\mu_o =17.5$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test if the true mean is higher than 17.5, the system of hypothesis would be:

Null hypothesis: $\mu \leq 17.5$

Alternative hypothesis: $\mu > 17.5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing we got:

$t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$

The degrees of freedom are given by:

$df=n-1=45-1=44$

Part b

The p value would be given by:

$p_v =P(t_{(44)}>2.744)=0.0044$

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3 years ago

An office has 30 computers. Seventeen of the 30 are Macintosh, and the remaining thirteen are windows. Two computers are randoml

Alex Ar [27]

Answer:

0.5085

Step-by-step explanation:

Given that an office has 30 computers. Seventeen of the 30 are Macintosh, and the remaining thirteen are windows.

When two computers are randomly selected without replacement

no of ways of selection = $60C2 = 1770$

No of ways of selecting one window and one Mac = $30C1(30C1) = 900$

Prob that the sample contains exactly one windows machine and exactly one Macintosh= $\frac{900}{1770} =0.5085$

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3 years ago

The mean age when smokers first start is 13 years old with a population standard deviation of 2 years. A researcher thinks that

lakkis [162]

Answer:

Null hypothesis: $\mu = 13$

Alternative hypothesis: $\mu \neq 13$

$z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191$

a. Reject the null hypothesis

c. There is enough evidence to support the claim

Step-by-step explanation:

Data given and notation

$\bar X=12/2$ represent the sample mean

$s=20000$ represent the standard deviation for the sample

$\sigma =2$ represent the population standard deviation

$n=30$ sample size

$\mu_o =13$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the smoking age has significantly changed since the invention of ENDS—electronic nicotine delivery systems, the system of hypothesis would be:

Null hypothesis: $\mu = 13$

Alternative hypothesis: $\mu \neq 13$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191$

What do you conclude? Using the p-value approach

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

$p_v =2*P(Z$

Since the $p_v$ then the correct decision is:

a. Reject the null hypothesis

And the best conclusion is :

c. There is enough evidence to support the claim

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3 years ago