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Alisiya [41]
2 years ago
6

given the arithemtic or geometric sequence, identify f(0) and the common difference ratio f(x) = 4-2/3x

Mathematics
1 answer:
ankoles [38]2 years ago
7 0

<u>Answer: </u>

Required value of f(0) = 4 and common difference of given sequence is \frac{-2}{3}

<u>Solution: </u>

Given sequence rule for arithmetic sequence is  \mathrm{f}(\mathrm{x})=4-\left(\frac{2}{3}\right) x

We need to determine f(0) and common difference

Calculating f(0).

Substituting x = 0 in given sequence rule  we get

\begin{array}{l}{f(0)=4-\left(\frac{2}{3}\right) 0} \\ {=4-0=4}\end{array}

Calculating common difference  

Let’s first calculate f(1) and f(2).

Substituting x = 1 in given sequence rule  

\begin{array}{l}{\mathrm{f}(1)=4-\left(\frac{2}{3}\right) 1} \\ {=4-\left(\frac{2}{3}\right)=\frac{10}{3}}\end{array}

\mathrm{f}(2)=4-\left(\frac{2}{3}\right) 2=4-\left(\frac{4}{3}\right)=\frac{8}{3}

Common Difference can be found by subtracting f(1) from f(2)

\text { Common difference }=\mathrm{f}(2)-\mathrm{f}(1)=\left(\frac{8}{3}\right)-\left(\frac{10}{3}\right)=\frac{-2}{3}

Hence required value of f(0) = 4 and common difference of given sequence is \frac{-2}{3}

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Answer:

The probability is 0.0428

Step-by-step explanation:

First, let's remember that the binomial distribution is given by the formula:

P(X=k) =\left[\begin{array}{ccc}n\\k\end{array}\right] p^{k}(1-p)^{n-k} where k is the number of successes in n trials and p is the probability of success.

However, the problem tells us that when there isn't a number of trials fixed, we can use the geometric distribution and the formula for getting the first success on the xth trial becomes:

P(X=x) = p(1-p)^{x-1}\\

The problem asks us to find the probability of the first success on the 4th trial (given that the first subject to be a universal blood donor will be the fourth person selected)

Using this formula with the parameters given, we have:

p = 0.05

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Substituting these parameters in the formula and solving it, we get:

P(X=4) = 0.05(1-0.05)^{4-1}\\P(X=4) = 0.05 (0.95)^{3}\\P(X=4) = 0.05(.8573)\\P(X=4) = 0.0428

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7 0
3 years ago
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

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

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

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Please put more details in the question 
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299,792,458 rounded to the greatest place value
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Answer:

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Step-by-step explanation:

299,792,458

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