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3 years ago

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Mathematics

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erastova [34]3 years ago

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Find the interquartile range (IQR) of the data in the box plot below. A horizontal boxplot titled Magnitude of earthquakes, is p

tresset_1 [31]

Answer:

IQR = 3

Step-by-step explanation:

Interquartile range, which is abbreviated as IQR, is just the difference between the upper quartile (where the box ends) and the lower quartile (where the box starts). AS it is given in the question that that box starts from 4 and ends at . The interquartile range IQR is simply:

IQR = 7 - 4 = 3

The values of data are usually clustured around some central value. Interquartile range tells us how spread out these middle values. It also tells when some values are too far from the central value. The values which are too far are also termed as outlier values

8 0

3 years ago

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A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n

Olegator [25]

Answer:

i) $2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959$

$2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041$

So then the 90% confidence interval is given by (1.959, 2.041)

ii) Figure attached

iii) $ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041$

Step-by-step explanation:

Previous concepts

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".

The confidence interval is given by this formula:

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

And for a 90% of confidence the significance is given by $\alpha=1-0.9=0.1$ , and $\frac{\alpha}{2}=0.05$ . Since we know the population standard deviation we can calculate the critical value $z_{0.05}= \pm 1.64$

We know the folllowing data:

$\bar X = 2$ represent the sample mean

$\sigma = 0.1$ represent the population deviation

n =16 represent the sample size

Part i)

If we replace the values given into formula (1) we got:

$2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959$

$2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041$

So then the 90% confidence interval is given by (1.959, 2.041)

Part ii)

Figure attached. We have the illustration for the confidence interval obtained.

Part iii)

The margin of error is given by:

$ME=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}$

And if we replace we got:

$ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041$

3 0

4 years ago

6sacks of potatoes weigh 122.4 kg and 4 sacks of onions weigh 50 kg. What is the total amount of 1 sack of potatoes and 1 sack o

lys-0071 [83]

Answer: See explanation

Step-by-step explanation:

6 sacks of potatoes weigh 122.4 kg

1 sack would weigh 122.4 kg ÷ 6

122.4 ÷ 6 = 20.4

1 sack of potatoes weighs 20.4 kg

4 sacks of onions weigh 50 kg

1 sack would weigh 50 kg ÷ 4

50 ÷ 4 = 12.5

1 sack of onions weighs 12.5 kg

4 0

3 years ago

I need to know if this number is any of the following:

Nikitich [7]

It is rational/irrational

8 0

3 years ago

Read 2 more answers

The ratio of elena's height to pavai's height is 10:13. The ratio of elena's height to kamir's height is 2:3. Their total height

tensa zangetsu [6.8K]

Answer:

a) 10:13:3

b) 258 cm

Step-by-step explanation:

The ratio of elena's height to pavai's height is 10:13. The ratio of elena's height to kamir's height is 2:3. Their total height is 456 cm.

a) Find the ratio of elena's height to pavani's height to kamir's height in its simplest form.

10:13:3

b) What is pavani's height?

The ratio of elena's height to pavai's height is 10:13.

Their total height is 456 cm

Sum of proportions = 10 + 13 = 23

Hence:

13/23 × 456 cm

= 257.73913043 cm

Approximately = 258 cm

3 0

3 years ago

The picture is the question​

The picture is the question