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Ierofanga [76]
3 years ago
10

Select the correct answer.

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

ypur answer would be a.

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Which answer is the correct form of bar notation for this decimal number: 4.676767…
Rudiy27
Hello There!

Well all of your answer options are the same so i suppose they are all right.

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Read 2 more answers
BRAINLIEST AWARD NO.2
zzz [600]

Answer:

A whole number first term to render as fifth term a value larger than 10000, should be at least 121

Step-by-step explanation:

The formula is given as recursive since it involves the previous number of the sequence, and defined as:

a_n=a_{n-1}*3+6

we also know that the first term is 4

Then in this case, the first five terms are:

a_1=4\\a_2=4*3+6=18\\a_3=18*3+6=60\\a_4=60*3+6=186\\a_5=186*3+6=564\\

So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2

So the rule is to subtract 6 from he term, and divide the subtraction by 3. Then working backwards:

a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49

therefore, the starting first term should be at least about 121 to give a fifth term larger than 10,000

8 0
3 years ago
It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele
kow [346]

Answer:

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

\bar X=23500 represent the sample mean  

s=3900 represent the sample standard deviation      

n=100 sample size      

\mu_o =2000 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t 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.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:\mu \leq 2000      

Alternative hypothesis:\mu > 2000      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

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

t-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".  

Calculate the statistic      

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

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

df=n-1=100-1=99  

Since is a one-side upper test the p value would be:      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

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