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3 years ago

Tom made the box plots to compare the number of calories between his morning snacks and his afternoon snacks.

Mathematics

1 answer:

serg [7]3 years ago

3 0

Answer:

cc ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Step-by-step explanation:

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The table below shows the average temperature, in degrees Celsius, in Jacobs city over a period of five months

katen-ka-za [31]

Linearly, because the table shows that temperature increased by the same amount each month (2.2). If it were exponentially, the temperature would increase by the same percentage each month.

3 0

3 years ago

At Emily’s closet 1/3 of the cloths are white and 1/5 are black what. What fraction of Emilly’s clothes are black and white ?

vfiekz [6]

8/15

Since 1/5 and 1/3 have different denominators, they cannot be added directly.
Multiply the denominators (5×3=15)
Multiply each with the other denominator 3×(1/5)=3/15, 5×(1/3)=5/15
Now, 3/15 +5/15=8/15

3 0

4 years ago

Read 2 more answers

Researchers are interested in the effect of a certain nutrient on the growth rate of plant seeding. Use a hydroponics grow proce

Rashid [163]

Answer:

The 95% confidence interval for the mean would be given by (56.604;68.596)

The 99% confidence interval for the mean would be given by (53.196;72.004)

Step-by-step explanation:

1) Previous concepts

When we compute a confidence interval for the mean, we are interested on the parameter population mean, and we use the info from the sample to estimate this parameter.

2) Basic operations

The sample mean can be calculated with the following formula

$\sum_{i=1}^n \frac{x_i}{n}$

Using excel we can use this function to calculate the mean:

=AVERAGE(54.2,59.8,61.8,63.3,65.1,71.4)

The value obtained is $\bar X=62.6$

In order to find the sample deviation we can use this formula

$s=\sqrt{\sum_{i=1}^n \frac{(x_i-\bar X)^2}{n-1}}$

And using excel we can use this function to calculate the sample standard deviation:

=STDEV.S(54.2,59.8,61.8,63.3,65.1,71.4)

The value obtained is $s=5.713$

The sample size for this case is n=6, n<30 so then is better use the t distribution to calculate the margin of error. First we need to calculate the degrees of freedom, on this case $df=n-1=6-1=5$

The formula for the confidence interval would be given by:

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

3) Part a

If we want a 95% or 0.95 of confidence, then the value for the signficance is $\alpha=1-0.95=0.05$ , and $\alpha/2=0.025$ , and $1-\frac{\alpha}{2}=0.975$ so we can find the critical t value with the following formula in excel:

=T.INV(0.975,5)

And we got $t_{\alpha/2}= 2.571$

And we can replace into equation (1) and we got:

$62.6 \pm 2.571\frac{5.713}{\sqrt{6}}$

And using excel with the following formulas we got:

=62.6-2.571*(5.713/SQRT(6)) = 56.604

=62.6+2.571*(5.713/SQRT(6)) = 68.596

So the 95% confidence interval for the mean would be given by (56.604;68.596)

4) Part b

If we want a 99% or 0.99 of confidence, then the value for the signficance is $\alpha=1-0.99=0.01$ , and $\alpha/2=0.005$ , and $1-\frac{\alpha}{2}=0.995$ so we can find the critical t value with the following formula in excel: