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4vir4ik [10]
3 years ago
12

The side of a square measures (2x − 5) units. Part A: What is the expression that represents the area of the square? Show your w

ork to receive full credit. (4 points) Part B: What are the degree and classification of the expression obtained in Part A? (3 points) Part C: How does Part A demonstrate the closure property for polynomials? (3 points)
Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

A. The area of a square is given as:

<span>A = s^2 </span>

Where s is a measure of a side of a square. s = (2 x – 5) therefore,

<span>A = (2 x – 5)^2                   </span>

Expanding,

A = 4 x^2 – 20 x + 25

 

<span>B. The degree of a polynomial is the highest exponent of the variable x, in this case 2. Therefore the expression obtained in part A is of 2nd degree.</span>

Furthermore, polynomials are classified according to the number of terms in the expression. There are 3 terms in the expression therefore it is classified as a trinomial.

 

<span>C. The closure property demonstrates that during multiplication or division, the coefficients and power of the variables are affected while during multiplication or division, only the coefficients are affected while the power remain the same.</span>

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In the flowering plant, white flowers (B) are dominant over red flowers (b), and short plants (E) are dominant over tall flower
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<u>Question Completion</u>

(a)What is your null hypothesis?

(b)What is your expected phenotypic ratio based on Mendelian inheritance?

(c)Calculate the expected number of flowers you should have gotten based on the Mendelian inheritance. Then calculate a chi-square value, degrees of freedom, and a p-value.

  • Chi-square statistic: _____
  • Degrees of freedom (# phenotypes -1):
  • P-value:

(d)Interpret your results. Do you reject it or fail to reject your null hypothesis (please restate the null)?

Answer:

(a)H_0:$The given data fit the predicted phenotype

(b)9:3:3:1

(c)

  • Chi-square statistic: 3.8914
  • Degrees of freedom (# phenotypes -1) =3
  • P-value:  0.2734

(d)We fail to reject the null hypothesis.

Step-by-step explanation:

In the flowering plant, white flowers (B) are dominant over red flowers (b), and short plants (E) are dominant over tall flowers (e). An F2 generation was created by crossing two F1 individuals (each BbEe).

(a)The null hypothesis is:

H_0:$The given data fit the predicted phenotype

(b)The gametes are BE, Be, bE and be.

The offsprings are presented in the table below:

\left|\begin{array}{c|cccc}&BE&Be&bE&be\\--&--&--&--&--\\BE&BE&BE&BE&BE\\Be&BE&Be&BE&Be\\bE&BE&BE&bE&bE\\be&BE&Be&bE&be\end{array}\right|

The expected phenotypic ratio based on Mendelian inheritance

BE:Be:bE:be=9:3:3:1

(c)

\left|\begin{array}{c|c|c|c|c|c}$Phenotype&Observed&$Expected&O-E&(O-E)^2&\dfrac{(O-E)^2}{E} \\-----&--&--&--&--&--\\$White short(BE)&206&\frac{9}{16}*404 \approx 227 &-21&441&1.9427\\$Red, short(bE)&83&\frac{3}{16}*404 \approx 78 &5&25&0.3205\\$White, tall(Be)&85&\frac{3}{16}*404 \approx 78 &7&49&0.6282\\$Red, tall(be)&30&\frac{1}{16}*404 \approx 25 &5&25&1\\-----&--&--&--&--&--\\$Total&404&--&--&--&3.8914\end{array}\right|

Therefore:

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We, therefore, fail to reject the null hypothesis.

The difference in the observed and expected are sosmall that they can be attributed to random chance.

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Hi there

Replace the blank space by x

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