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1 year ago

The major pathway of communication between the right and left hemispheres is the:.

Biology

1 answer:

Sergeu [11.5K]1 year ago

6 0

The major pathway of communication between the right and left hemisphere is the corpus callosum.

The cerebrum is the recognizable gray covering of the human brain, appearing to comprise the majority of its mass.
The cerebral cortex is the wrinkly area; the remainder of the anatomy is hidden behind that layer.
The longitudinal fissure, which is a significant gap between the two sides of the cerebrum, exists.
It divides the brain into right and left cerebral hemispheres, which are two independent parts.
The main channel for communication between the two hemispheres of the cerebral cortex is provided by the white matter of the corpus callosum, which is located deep inside the cerebrum.

learn more about corpus callosum here: brainly.com/question/13094825

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The coefficients in a balanced chemical equation always can express the ratio of

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The correct answer is D

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3 years ago

A 0.2 M solution of NaCl contains ________ g of NaCl in 1 l of ________.

Maksim231197 [3]

Answer:

11.7 g in 1 L of water.

Explanation:

Molarity (M) or Molar concentration is a measure used in chemistry for solutes in a solution.

It is defined as mol/ L (number of moles per litre).

0,2 M means we have 0,2 mol/ L

In order to know how many grams this means we need to look up the molecular weight of the solute. In this case NaCl: 58.443 g/mol

$\frac{1 mol NaCl}{58.443 g}$ = $\frac{0.2 mol}{X g}$

X= $\frac{(0.2 mol x 58.433 g}{1 mol}$

X= 11.69 g

The solution has 11.69 g of NaCl in 1 L of water since its an adequate polar solvent for NaCl dissolution.

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3 years ago

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.

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