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Neporo4naja [7]
3 years ago
13

Let f(x) = 4x − 3. If f(a) = 9 and f(b) = 5, then what is f(a + b) ?

Mathematics
1 answer:
kvv77 [185]3 years ago
7 0
F(x) = 4x - 3

f(a) = 4a-3

9 = 4a-3

4a = 12

a = 12 / 4

a = 3

_____________________

f(b) = 4b-3

5+3 = 4b

8 = 4b

b = 8 / 4

b = 2

___________________

f(a+b) = 4(a+b) - 3

f(a+b) = 4(3+2) - 3

f(a+b) = 4(5) - 3

f(a+b) = 20 - 3

f(a+b) = 17
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TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin
choli [55]

Answer:

a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51  

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation  

n=2341 represent the random sample taken    

X represent the people that they have watched digitally streamed TV programming on some type of device

\hat p=0.55 estimated proportion of people that they have watched digitally streamed TV programming on some type of device  

\alpha=0.01 represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval  

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

p \sim N (p, \sqrt{\frac{p(1-p)}{n}}

Part a) Confidence interval

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.55 - 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.523

0.55 + 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.577

And the 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51  

And rounded up we have that n=1831

8 0
4 years ago
For an independent-measures ANOVA comparing two treatments with a sample of n = 10 in each treatment, what is the critical value
trapecia [35]

Answer:

critical value is 4.41

Step-by-step explanation:

Given data

n = 10

α = .05

to find out

the critical value

solution

first we calculate degree of freedom i.e

degree of freedom = n -1

degree of freedom = 10 - 1

degree of freedom = 9

and

degree of freedom 2 =( n -1 ) 2

degree of freedom 2 =( 10 -1 ) 2

degree of freedom 2 = 18

so for degree of freedom 9 and degree of freedom 2 is 18  and α = .05

critical value is 4.41

4 0
3 years ago
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