PLEASE HELP!!! w(:_;)w(/´Д｀)/

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

2 years ago

A barrier separating the cell’s genetic information from the cell’s cytosol.

malfutka [58]

Answer:

All cells hace a barrier around them that separates them from the environment and from other cell.The barrier is called the plasma membrane,or cell membrane.

8 0

1 year ago

alguien que me ayude con esta pregunta por favor cuáles son las enfermedades que se desarrollan lentamente pudiendo durar mucho

Anarel [89]

Answer:

Enfermedad que se desarrolla lentamente, pudiendo durar mucho tiempo o incluso el resto de la vida de una persona. Un ejemplo es la artritis.

Explanation:

Tipo de glóbulo blanco o leucocito que fabrica unas sustancias llamadas anticuerpos, muy importantes en nuestras defensas.

Enfermedades que llevan a la muerte

Las diez enfermedades más misteriosas

Sida.

Alzhéimer.

La causa del alzhéimer todavía es desconocida. Las investigaciones suelen asociar la enfermedad con la aparición de placas seniles y ovillos neurofibrilares. ...

Resfriado común.

Gripe aviar.

Pica.

Enfermedades autoinmunes.

Esquizofrenia.

8 0

2 years ago

Which of the following is true of conifers?

cricket20 [7]

"<span>Female cones produce pollen that is trapped by the male cones’ scales" is the one among the following choices given in the question that is true of conifers. The correct option among all the options that are given in the question is the fourth option or the last option. I hope the answer helps you.</span>

7 0

2 years ago

Read 2 more answers

If phosphofructokinase experienced a mutation that interfered with substrate binding, then what other enzyme is going to be most

frozen [14]

Answer: Aldolase

Explanation:

In the metabolism of glucose( glycolysis) phosphofructokinase is an enzyme that catalyzes the conversation of fructose-6-phosphate into fructose-1,6-bisphosphate. This in turn is converted to pyruvate after various steps of enzymatic activity in the glycolytic pathway.

If phosphofructokinase experienced a mutation that interfered with substrate binding, the enzyme that is going to be most immediately impacted in terms of accessing substrate is the ALDOLASE.

Aldolase enzymes cleave fructose-1,6-bisphosphate to triose phosphates( glyceraldehyde 3-phosphate and dihydroxy-acetone phosphate) facilitating an increase in anaerobic production of ATP in muscle.

Therefore, the substrate for binding of aldolase, which is fructose-1,6-bisphosphate is lacking due to mutation of phosphofructokinase enzyme.

7 0

2 years ago