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3 years ago

A cell membrane is very specific about what it allows across how much does this help the cell

Biology

2 answers:

Lostsunrise [7]3 years ago

7 0

Answer:

Transport proteins allows only enzymes to pass through the membrane

Explanation:

givi [52]3 years ago

4 0

Answer:

It is the better for the cell if most molecules are prevented from entering.

Explanation:

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The class of macromolecule that would most likely be involved in contracting muscles would be___

Ksju [112]

Answer:

The correct answer is option - B. proteins.

Explanation:

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In the contraction, these protein filaments help to change in length and shape of the muscles. Thin filaments involve actin and thick filament in is myosin filaments.

Thus, the correct answer is option - b. proteins.

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3 years ago

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Dima020 [189]

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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