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grandymaker [24]
3 years ago
14

ANSWER QUICKLY PLS!! Select the graph that best illustrates the story.

Mathematics
1 answer:
Andrej [43]3 years ago
5 0

Answer:

C

Step-by-step explanation:

We start too shrink slowly or just stop growing at a certain age

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70

Step-by-step explanation:

1 times 70 = 70

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A carpenter is installing a circular light fixture with a diameter of 6 inches in the ceiling. To do this, she must cut a square
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3 years ago
Marshal compares water tanks at home improvment store. One tank holds 2.02 gallons ofof water a second tank holds 28.1 gallons o
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28.1 - 2.02 = 26.08


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5 0
3 years ago
Write an equation of a parabola that opens to the left, has a vertex at the origin, and a focus at (–9, 0.
Georgia [21]
The standard equation of parabola:

(y-k)²=4p(x-h), with:

a) vertex = (h,k)

b) focus = (h+p, k)

c) directrix = (x=h-p)


Since this parabola has a vertex at (0,0) that means h=k=0

Hence the equation becomes: y²=4px, let's calculate p:
focus is given (-9,0) Remember h+p = -9 & since h=0, then p= -9
===> y²= - 36x
8 0
4 years ago
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
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