Which condition is a malignant tumor composed of blood-forming tissues of the bone marrow?

Question 9.9. Which bone would a forensic anthropologist analyze to identify a victim as male or female?

OLEGan [10]

The answers would be:

Pelvis

Coroner's office

Here is more about your questions:

The pelvis is a good identifier when it comes to male and female because the features have distinct features. The female pelvis for example is broader and the inlet is much wider. It is structured that way to prepare the woman's body for pregnancy and child birth.

Forensic anthropologists is a sub-category of anthropology. Forensic anthropologists examine skeletal remains to determine how an individual lived, died, and how long they were dead. They can also identify who they were. Forensic anthropologists can assist especially when the remains are badly decomposed.

5 0

3 years ago

Read 2 more answers

Click the statement that best describes the role of each element in living organisms.

podryga [215]

Nitrogen should be the answer

5 0

3 years ago

Read 2 more answers

Select ALL the correct answers.

tekilochka [14]

Answer:

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Explanation:

Given

$A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}$

$10th\ run = 9$

So:

$B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}$

Required

Select all true statements

(a) & (d) Median Comparisons

$A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}$ $B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}$

$n = 9$ $n = 10$

Arrange the data:

$A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}$ $B = \{1,1,1,1,2,2,2,3,4,9\}$

$Median = \frac{n + 1}{2}th$

$Median = \frac{9 + 1}{2}th$ $Median = \frac{10 + 1}{2}th$

$Median = \frac{10}{2}th$ $Median = \frac{11}{2}th$

$Median = 5th$ $Median = 5.5}th$ --- average of 5th and 6th

$Median = 2$ $Median = \frac{2+2}{2} = 2$

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

$A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}$ $B = \{1,1,1,1,2,2,2,3,4,9\}$

$n = 9$ $n = 10$

First, calculate the lower quartile (Q1)

$Q_1 = \frac{n + 1}{4}th$ [Odd n] $Q_1 = \frac{n}{4}th$ [Even n]

$Q_1 = \frac{9 + 1}{4}th$ $Q_1 = \frac{10}{4}th$

$Q_1 = \frac{10}{4}th$ $Q_1 = 2.5$

$Q_1 = 2.5th$

This means that:

$Q_1 = 2nd + 0.5(3rd - 2nd)$ $Q_1 = 2nd + 0.5(3rd - 2nd)$

$Q_1 = 1 + 0.5(1- 1)$ $Q_1 = 1+ 0.5(1 - 1)$

$Q_1 = 1$ $Q_1 = 1$

Next, calculate the upper quartile (Q3)

$Q_3 = \frac{3}{4}(n + 1)th$ [Odd n] $Q_3 = \frac{3}{4}(n)th$ [Even n]

$Q_3 = \frac{3}{4}(9 + 1)th$ $Q_3 = \frac{30}{4}th$

$Q_3 = \frac{30}{4}th$ $Q_3 = 7.5th$

$Q_3 = 7.5th$

This means that:

$Q_3 = 7th + 0.5(8th- 7th)$ $Q_3 = 7th + 0.5(8th- 7th)$

$Q_3 = 2 + 0.5(3- 2)$ $Q_3 = 2+ 0.5(4 - 2)$

$Q_3 = 2.5$ $Q_3 = 3$

The interquartile range is $IQR = Q_3 - Q_1$

So, we have:

$IQR = 2.5 - 1$ $IQR = 3 - 1$

$IQR = 1.5$ $IQR =2$

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal

8 0

3 years ago

A particular leg bone for dinosaur fossils has a mean length of 60 inches with a standard deviation of 3 inches. What is the len

stepan [7]

The average length of a specific dinosaur leg bone is 60 inches, with a standard variation of 3 inches. If a z-score is 0, then the length is 67.

Divide your data point by your standard error after taking your data point and subtracting the mean from it. How is the z-score determined? Z-scores are calculated using the formula z = (x-)/, where x represents the raw score, the population mean, & the population variation. The z-score is just the raw score less the population mean, split by the population variance, as the calculation demonstrates. You now have your Z-score. For instance, a scoring of 450 would equate a scores of (450-500)/100 (= -50/100 = -0.50 on a scale with a mean of 500 as well as a standard variation of 100, indicating that the scoring is half a variance below the mean.

Learn more about variation

brainly.com/question/17287798

#SPJ4

7 0

1 year ago

In snapdragons, there is one allele that produces red flowers and another allele that produces white flowers. Neither allele is

AVprozaik [17]

You do:
R R
R
r
Your offspring would be: RR, RR, Rr, Rr
So, about 50% of the offspring will have red flowers
Your answer is
50%

5 0

3 years ago