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3 years ago

Why must s-phase occur before mitosis? ...?

1 answer:

6 0

Here is the answer of the given question above.
S phase happens after G1 phase and this is a crucial period because this is when DNA is copied. G2 phase then happens, which is prior to mitosis. Once this is complete, this then mitosis happens wherein the cells divides the contents of the nucleus between two daughter cells. Hope this answer helps.

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The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.

Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=5$ , we get,

$V=\frac{4}{3} (3.14)(125)$

$V=\frac{1580}{3}$

$V=523.3333 \ in^3$

Rounding off to two decimal places, we have,

$V=523.33 \ in^3$

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = $\frac{3}{4} (5)=3.75$

Radius of the third sphere = $\frac{3.75}{2} =1.875$

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=1.875$ , we get,

$V=\frac{4}{3} (3.14)(1.875)^3$

$V=\frac{4}{3} (3.14)(6.59)$

$V=27.5901 \ in^3$

Rounding off to two decimal places, we have,

$V=27.59 \ in^3$

Thus, the volume of the third sphere is 27.59 in³

4 0

4 years ago

!!!!!HELP ME PLZ!!!!!

Brilliant_brown [7]

Answer:

y=-2

Step-by-step explanation:

in other words, the question is asking

"six time what plus four equals 12?"

we know that 6 ⋅ 2 = 12

therefore if we plug the -2 into the Y we get

6(-2+4)=12

6(2)=12 or 6⋅2=12

therefore the answer is y=-2

4 0

3 years ago

a cookie recipe requires 8 tablespoons of butter. If this recipe is used to bake 30 cookies, how much butter is needed for 75 co

blsea [12.9K]

Answer:

9.375

Step-by-step explanation:

3 0

3 years ago

In GHJ, GH = 21 inches, HJ = 24 inches, and GJ = 9 inches. Which statement is true? A. G is smallest B. J is smallest C. H is la

Semmy [17]

Answer:

D. G is the largest

Step-by-step explanation:

1. Draw a triangle and label it so it says G, H, and J, on its points.

2. Write 21 between the G and H. Write 24 between the H and J. Write 9 between G and J.

3. Since side HJ is 24 inches, larger than GH and GJ, the angle across from it is also the largest. If you look at your drawn triangle the angle across from side HJ is angle G.

5 0

3 years ago

Employees at a company produced refrigerators on three shifts. Each shift recorded their quality stats below. A unit was conside

Lady bird [3.3K]

Answer:

Step-by-step explanation:

Hello!

So in the refrigerator factory there are three shifts. Each shift records their quality based on the quantity of defective and working parts assembled.

Using a Chi-Square test of independence you have to test the claim that quality and shifts are independent.

The hypotheses are:

H₀: The variables are independent.

H₁: The variables are not independent.

α: 0.05

$X^2= sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~X_{(r-1)(c-1)}$

r= total number of rows

c= total number of columns

i= 1, 2 (categories in rows)

j=1, 2, 3 (categories in columns)

To calculate the statistic you have to calculate the expected frequencies for each category:

$E_{ij}= \frac{O_{i.}*O_{.j}}{n}$

$O_{i.}$ Represents the marginal value of the i-row

$O_{.j}$ Represents the marginal value of the j-column

$E_{11}= \frac{O_{1.}*O_{.1}}{n}= \frac{21*40}{120}= 7$

$E_{12}= \frac{O_{1.}*O_{.2}}{n}= \frac{21*40}{120}= 7$

$E_{13}= \frac{O_{1.}*O_{.3}}{n}= \frac{21*40}{120}= 7$

$E_{21}= \frac{O_{2.}*O_{.1}}{n}= \frac{99*40}{120}= 33$

$E_{22}= \frac{O_{2.}*O_{.2}}{n}= \frac{99*40}{120}= 33$

$E_{23}= \frac{O_{2.}*O_{.3}}{n}= \frac{99*40}{120}= 33$

$X^2_{H_0}= \frac{(7-7)^2}{7} + \frac{(5-7)^2}{7} + \frac{(9-7)^2}{7} + \frac{(33-33)^2}{33} + \frac{(35-33)^2}{33} + \frac{(31-33)^2}{33} = 1.385= 1.34$

Using the critical value approach, the rejection region for this test is one-tailed to the right, the critical value is:

$X^2_{(c-1)(r-1);1-\alpha }= X^2_{2; 0.95}= 5.991$

Decision rule:

If $X^2_{H_0}$ ≥ 5.991, reject the null hypothesis.

If $X^2_{H_0}$ < 5.991, do not reject the null hypothesis.

The value of the statistic is less than the critical value, the decision is to not reject the null hypothesis.

At 5% significance level, you can conclude that the shift the pieces were assembled and the quality of said pieces are independent.

I hope this helps!

4 0

3 years ago