A human brain weighs about 1 kg and contains about 1011 cells. Assuming that each cell is completely filled with water (density = 1 g/mL), calculate the length of one side of such a cell if it were a cube. If the cells were spread out into a thin layer that was a single cell thick, what would be the total surface area (in square meters) for one side of the cell layer?

Answers

Answer 1
Answer:

Answer:

Step-by-step explanation:

From the information given:

a human brain weighs = 1 kg ; = 1000 grams

Number of cells = 10¹¹ cells

The density of water filled in each cell = 1 g/mL

From above;

the weight of each of the brain cell = total weight of the human brain/the number of cells

the weight of each of the brain cell = 1000/10¹¹

the weight of each of the brain cell = 1 × 10⁻⁸ grams

Now, to calculate the quantity of water in each cell; we have:

= the weight of each brain × density

= 1 * 10^(-8) \ g * (1 \ mL)/(1   \ g)= 1 * 10^(-8) \ mL

For cube; we know that

1 mL = 1 cm³

Thus:

1 * 10^(-8) \ mL=  1 * 10^(-8) \ cm^3

Recall that; the volume of a cube as well = x^3

where;

x = length of each sides

x^3 = 1 * 10^(-8) \ cm^3

x = \sqrt[3]{1 * 10^(-8)}

x = 0.0022 cm

Thus, the length of each side of the cell = 0.0022 cm

The surface area of a single cell = x²

The surface area of a single cell = (0.0022 cm)²

The surface area of a single cell = 4.84 × 10⁻⁶ cm²

Therefore, the total surface area of  is:

1 * 10^(11)  \ cells * (4.84 * 10^(-6) \ cm^2)/(1 \ cell)

= 4.84 * 10^5 cm^2

= 5.0* 10^5 cm^2

= 50 m²

Answer 2
Answer:

Final answer:

If the human brain's cells were cube-shaped and filled with water, each cell would be roughly 21.5 micrometers on a side. If these cells were spread out into a single-cell-thick layer, the total surface area for one side of the layer would be approximately 4.63 square meters.

Explanation:

To answer your question, the human brain has about 1011 cells, each filled with water. Given the total mass of the brain (about 1 kg) and the number of cells, we can calculate the volume of a single cell. The density of water is 1 g/mL or 1,000 kg/m³, so the volume of all the cells (entire brain) is 1 m³. Therefore, the volume of a single cell must be 1 m³/1011 cells, which is approximately 10-14 m³. For a cubical cell, the side length of the cube (a) would be the cube root of this volume, which is approximately 2.15 x 10-5 m or 21.5 micrometers.

To calculate the total surface area for one side of the cell layer, we multiply the area of a single cell by the total number of cells: (2.15 x 10-5)² m²/cell x 1011 cells = approximately 4.63 m².

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I think it’s but I’m not sure E

Answer:

:)

Step-by-step explanation:

Find the sum of the two polynomials (2x7−5x6−7x5+8)+(9x7−2x6−2x5−7)

Answers

Here you go! Hope this helps

I really need help with #1 a and b

Answers

1a. Gabriella uses these services:
  Checking (she has an account)
  Debit card


1b. At A+ Bank, Gabriella's monthly charges will be
  $1 (for checking) + $2.50 (for debit card) = $3.50

At NextGen Bank, Gabriella's monthly charges will be
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1c. A+ Bank is a better deal for Gabriella. The fixed fees there are lower than the variable fee at NextGen Bank because she uses her debit card so much.

This exercise involves the formula for the area of a circular sector. the area of a sector of a circle with a central angle of 170° is 70 m2. find the radius of the circle. (round your answer to one decimal place.) m

Answers

The formula of interest is
A = (1/2)r²·θ
You have A = 70 m², θ = 170° = 17π/18 rad. Then the radius is found from
70 m² = (1/2)r²(17π/18)
(36·70 m²)/(17π) = r²
r ≈ √47.18476 m
r ≈ 6.9 m

The radius of the circle is 6.9 m.

How to find the radius of the circle?

The  area of the sector of a circle given by the formula:

A = /360 *  (r²)

where r is the radius of the circle and is the central angle of the sector

We have:

= 170°

A = 70 m²

We need to solve for r:

A = /360 *  (r²)

70 = 170/360 * * r²

r² = (70 * 360)/(170 * )

r² = 47.18

r = √47.18

r = 6.9 m

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Heavy children: Are children heavier now than they were in the past? The National Health and Nutrition Examination Survey (NHANES) taken between 1999 and 2002 reported that the mean weight of six-year-old girls in the United States was 49.3 pounds. Another NHANES survey, published in 2008, reported that a sample of 190 six-year-old girls weighed between 2003 and 2006 had an average weight of 46 pounds. Assume the population standard deviation is =σ17 pounds. Can you conclude that the mean weight of six-year-old girls in 2006 is different from what it was in 2002? Use the =α0.10 level of significance and the critical value method.

Answers

Answer:

Step-by-step explanation:

Hello!

You have two surveys that measure the weight of six-year-old girls in the USA,

1) 1999-2002

μ= 49.3 pounds

(I'll take this mean as the population value since it can be considered "historical data" or point of comparison to make the test.)

2)2003-2006

sample n= 190

sample mean x[bar]= 46 pounds

population standard deviation σ= 17 pounds

Assuming that the study variable X" Weight of six-year-old girls between 2003 - 2006" (pound) has a normal distribution.

If you need to test that the children are heavier now (2003-2006) than in the past (1999-2002) the test hypothesis is:

H₀: μ ≤ 49.3

H₁: μ > 49.3

α: 0.10

The statistic is Z= (x[bar]-μ)   ~N(0;1)

                                (δ/√n)

The critical region is one-tailed to the right.

Z_(1-\alpha ) = Z_(0.90) = 1.28

So you'll reject the null hypothesis if the calculated statistic is equal or greater than 1.28.

Z=  46 - 49.3 = -2.67

     17/√190

Since the calculated value -2.67 is less than 1.28 you do not reject the null hypothesis. In other words, the six-year-old girls from 2003-2006 are thinner than the girls from 1999-2002.

I hope it helps!

Jenna flips two pennies 105 times. How many times can she expect both coins to come up heads?

Answers

The number of times that both coins come up heads will be 26.25.

What is the expected value?

In parameter estimation, the expected value is an application of the weighted sum. Informally, the expected value is the simple average of a considerable number of individually determined outcomes of a randomly picked variable.

The expected value is given below.

E(x) = np

Where n is the number of samples and p is the probability.

If two coins are flipped. Then the total number of the event is given as,

Total = 2 x 2 = 4 {HH, HT, TH, TT}

Favorable event = 1 {HH}

The probability of getting both same, then we have

P = 1/4

Jenna flips two pennies 105 times. Then the number of times that both coins come up heads will be given as,

E(x) = p × n

E(x) = 1/4 × 105

E(x) = 26.25

The number of times that both coins come up heads will be 26.25.

More about the expected value link is given below.

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Hi there!

Since the chance of one coin landing on heads is 1/2, we should multiply.

1/2 × 1/2 = 1/4

105 × 1/4 = 26.25

So, the answer is 26.25 times.

Hope this helps!
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