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vlabodo [156]
3 years ago
6

Question 2 of 5

Mathematics
2 answers:
loris [4]3 years ago
7 0

Answer:

From Edmentum :)

AlekseyPX3 years ago
6 0

Given:

The different recursive formulae.

To find:

The explicit formulae for the given recursive formulae.

Solution:

The recursive formula of an arithmetic sequence is f(n)=f(n-1)+d, f(1)=a,n\geq 2 and the explicit formula is f(n)=a+(n-1)d, where a is the first term and d is the common difference.

The recursive formula of a geometric sequence is f(n)=rf(n-1), f(1)=a,n\geq 2 and the explicit formula is f(n)=ar^{n-1}, where a is the first term and r is the common ratio.

The first recursive formula is:

f(1)=5

f(n)=f(n-1)+5 for n\geq 2.

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 5. So, the explicit formula for this recursive formula is:

f(n)=5+(n-1)(5)

f(n)=5+5(n-1)

Therefore, the correct option is A, i.e., f(n)=5+5(n-1).

The second recursive formula is:

f(1)=5

f(n)=3f(n-1) for n\geq 2.

It is the recursive formula of a geometric sequence with first term 5 and common ratio 3. So, the explicit formula for this recursive formula is:

f(n)=5(3)^{n-1}

Therefore, the correct option is F, i.e., f(n)=5(3)^{n-1}.

The third recursive formula is:

f(1)=5

f(n)=f(n-1)+3 for n\geq 2.

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 3. So, the explicit formula for this recursive formula is:

f(n)=5+(n-1)(3)

f(n)=5+3(n-1)

Therefore, the correct option is D, i.e., f(n)=5+3(n-1).

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Here are summary statistics for randomly selected weights of newborn​ girls: nequals=174174​, x overbarxequals=30.930.9 ​hg, seq
MaRussiya [10]

Answer:

The 95% confidence interval would be given by (29.780;32.020)  

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=30.9 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=174-1=173

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that t_{\alpha/2}=1.97, this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.  

Now we have everything in order to replace into formula (1):

30.9-1.97\frac{7.5}{\sqrt{174}}=29.780    

30.9+1.97\frac{7.5}{\sqrt{174}}=32.020

So on this case the 95% confidence interval would be given by (29.780;32.020)    

The value 29.6 is not contained on the interval calculated.

5 0
3 years ago
Which expression shows -5y (4-2y) + 8y^2-3y + 4 in simplest form?​
Tcecarenko [31]

Simplified: 18y^2-20y+1

3 0
2 years ago
In an August 2012 Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), 536 said that they were dissatisfied w
soldier1979 [14.2K]

Answer:

Yes, result is significant ; PVALUE < α

Step-by-step explanation:

Given :

x = 536

n = sample size = 1012

Phat = x / n = 536 / 1012 = 0.5296 = 0.53

H0 : P0 = 0.5

H1 : P0 > 0.5

Test statistic :

(Phat - P0) ÷ sqrt[(P0(1 - P0)) / n]

1-P0 = 1 - 0.5 = 0.5

(0.53 - 0.5) ÷ sqrt[(0.5*0.5)/1012]

0.03 ÷ 0.0157173

= 1.9087

Pvalue :

Using the Pvalue from test statistic :

Pvalue = 0.02815

To test if result is significant :

α = 0.05

0.02815 < 0.05

Pvalue < α ; Hence, result is significant at α=0.05; Hence, we reject H0.

3 0
3 years ago
Need help can’t get this question wrong. Am I doing this right. And how would I write this. When I get the answer. Very confused
olya-2409 [2.1K]
Please show the question you’re confused about. We can’t help if we don’t know the problem :)
5 0
3 years ago
Please help Will give you the brainliest.
balu736 [363]

Answer: A.

Step-by-step explanation:

7 0
3 years ago
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