Which statement correctly describes the organization of cells, tissues, organs, and organ systems within a human body?

I don’t have a question because I posted a photoz

Scorpion4ik [409]

Part a:

$Q_{1}$ = 56

$Q_{2}$ = 60

$Q_{3}$ = 63

The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The $Q_{2}$ is the overall medium, $Q_{1}$ is the medium of the first half, and $Q_{3}$ is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between $Q_{1}$ and $Q_{3}$ . In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + $\frac{7}{2}$ ) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0

2 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³

Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, $\rho_o$ = 800 kg/m³

density of water, $\rho_w$ = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0

3 years ago

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

vitfil [10]

Answer:

0.20

Explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force: $F_f=\mu mg$ , backward, with

$\mu$ coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

$F-F_f = 0$

which we can solve to find the coefficient of kinetic friction:

$F-\mu mg=0\\\mu = \frac{F}{mg}=\frac{99 N}{(50 kg)(9.8 m/s^2)}=0.20$

7 0

3 years ago

When Cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.

Allisa [31]

Answer:

i have no clue im sorry

Explanation:

8 0

3 years ago