"Two haploid cells are formed" happens in meiosis during telophase I.
<u>Answer:</u> Option C
<u>Explanation:</u>
The chromosomes are inserted into nuclei during telophase I. The cell now needs to undergo a cytokinesis cycle, which separates the initial cell's cytoplasm into two daughter cells. One set of chromosomes are contained in each daughter cell and called as haploid or half the original cell's overall chromosomes number.
The parent cell has two poles, each with a full haploid set of chromosomes (consisting sister chromatids) when the meiotic process hits Telophase I. A cleavage furrow is formed at this point, splitting the cytoplasm of the cell into a half (cytokinesis). Once the cytoplasm is completely separated, the two originating daughter cells start planning for the second meiotic division.
Mangrove provides shelter to organisms on high coats, they help prevent erosion along coasts and they filter and process pollutants out of the environment
To generate
electricity .
Answer:
1.875 is the expected number of children the couple will have.
Explanation:
<em>The probability of an event </em><em>X</em><em> is denoted by </em><em>P(X)</em><em>.</em>
Let assume that X is the number of children to whom the couple will give birth. The couple can have at most <em>four children</em> with the following probabilities.
- The couple will have one child if the first is boy, thus:
For one child P(X)=1/2=0.50
- The couple will have two children if the first is girl and the second is a boy, thus:
For two children P(X)=0.5×0.5=0.25
- The couple will have three children if the first two are girls and the third is a boy, thus:
For three children P(X)= 0.5×0.5×0.5=0.125
- The couple will have four children if the first three are girls and the fourth is a boy, or if all the four are girls. thus:
For four children P(X) =(0.5×0.5×0.5×0.5)+(0.5×0.5×0.5×0.5)=0.125
<u>Now calculating the expected number of children the couple will have:</u>
NOTE: Formula is given in the attached pic
Expected number of children = E(X)= (1×0.50)+(2×0.25)+(3×0.125)+(4×0.125)
E(X)= 1.875