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2 years ago

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Relationship B has a lesser rate than Relationship A. The graph represents Relationship A. Which table could represent Relation

ship B? First-quadrant graph showing a ray from the origin through the points (2, 3) and (4, 6). Horizontal axis label is Time in weeks. Vertical axis label is Plant growth in inches.
A. Time (weeks) 3 6 8 10 Plant growth (in.) 2.25 4.5 6 7.5

B. Time (weeks) 3 6 8 10 Plant growth (in.) 4.8 9.6 12.8 16

C. Time (weeks) 3 4 6 9 Plant growth (in.) 5.4 7.2 10.8 16.2

D. Time (weeks) 3 4 6 9 Plant growth (in.) 6.3 8.4 12.6 18.9

2 answers:

anzhelika [568]2 years ago

6 0

Table A would be the best choice.

aivan3 [116]2 years ago

5 0

Table a
hope this helps xx

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A study was performed on green sea turtles inhabiting a certain area. Time-depth recorders were deployed on 6 of the 76 capture

polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean shell length = 51.3 cm

s = sample standard deviation = 6.6 cm

n = sample of turtles = 6

$\mu$ = true mean shell length

Now, the 99% confidence interval for $\mu$ = $\bar X \pm t_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$

Here, $\alpha$ = 1% so $(\frac{\alpha}{2})$ = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> $\mu$ = $51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }$

= [51.3 - 10.864 , 51.3 + 10.864]

= [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

3 0

2 years ago

How many distinct pairs of disjoint non-empty subsets of A are there, the union of which is all of A?

Mrac [35]

A = {1, 2, 5, 6, 8}
{1} U {2, 5, 6, 8}
{2} U {1, 5, 6, 8}
{5} U {1, 2, 6, 8}
{6} U {1, 2, 5, 8}
{8} U {1, 2, 5, 6}
{1, 2} U {5, 6, 8}
{1, 5} U {2, 6, 8}
{1, 6} U {2, 5, 8}
{1, 8} U {2, 5, 6}
{1, 2, 5} U {6, 8}
{1, 2, 6} U {5, 8}
{1, 2, 8} U {5, 6}
{1, 5, 6} U {2, 8}
{1, 5, 8} U {2, 6}
{1, 6, 8} U {2, 5}
The answer is 15 distinct pairs of disjoint non-empty subsets.

5 0

2 years ago

Minni is arranging 3 different music CDs in a row on a shelf. Create a sample space for the arrangement of a jazz CD (J), a pop

weeeeeb [17]

The sample space for the 3 different music CDs is as follows:
{JPR, JRP, PJR, PRJ, RJP, RPJ}

8 0

2 years ago

Read 2 more answers

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of

Thepotemich [5.8K]

<h2><u>Answer with explanation</u>:</h2>

As per given , we have

sample size : n= 65

degree of freedom : df=n-1=64

sample mean : $\overline{x}=19.5$

sample standard deviation : s= 5.2

Since , the population standard deviation is not given , so we apply t-test.

Significance level for 90% confidence : $\alpha=1-0.90=0.10$

t-critical value for significance level 0.10 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690$

Formula for Confidence interval :

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$

Then , 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.076$

$(19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)$

∴ 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.424, 20.576)

Significance level for 95% confidence : $\alpha=1-0.95=0.05$

t-critical value for significance level 0.05 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977$

Then , 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.288$

$(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)$

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)

4 0

2 years ago

Seven Applicants for a Job - F, G, H, I, J, K, and L are each to be interviewed on monday or Tuesday . Each

stiks02 [169]

Answer:

I would guess its D. Because it is the one that makes the most sense.

Step-by-step explanation:

H and L has to be on the same day with no other application so not A, B, or C because there is L in all of them.

Hope I helped!

6 0

1 year ago