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enot [183]
3 years ago
14

Caleb made a stack of popcorn balls on a table. He placed 16 on a table then added three more layers on top of them. Each layer

had four fewer balls than the previous layer. How many popcorn balls were in all in the stack
Mathematics
2 answers:
Bess [88]3 years ago
7 0
16 popcorn balls  1st layer                             16+12+8+4 = 40 popcorn balls in the
12 popcorn balls 2nd layer                                         stack.
8 popcorn balls 3rd layer
4 popcorn balls 4th layer
Basile [38]3 years ago
6 0
Hi! So if we lay out 16 popcorn balls, and each time we take away four, this is what would happen gradually,

16 bottom + 12 1st layer + 8 2nd layer + 4 3rd layer

16+12+8+4 = 40 popcorn balls!

( don't forget to use lables)
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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
lutik1710 [3]

<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:

  • Chi-square statistic: 3.8914
  • Degrees of freedom (# phenotypes -1):  4-1 =3
  • P-value:  0.2734

(d) Our null hypothesis is:

H_0:$The given data fit the predicted phenotype

Since p>0.05, the given data fit the predicted phenotypic ratio.

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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A dice is rolled 2 times. What is the probability of showing a 5 on every roll Round to 4 decimal places
AlexFokin [52]

Answer:

0.0278

Step-by-step explanation:

(1/6)(1/6) = 1/36 = 0.0278

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EXERCISE 3.4
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Answer:

I think it's C. sjisebd ddhisuaja. usually get ebe.

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1) Is 26,341 divisible by 3? If it is, write the number as the product of 3 and another factor. If not, explain.
yan [13]

Answer:

No

Step-by-step explanation:

26341 ÷ 3 = 8780.333333

because the product has a decimal of 8780 recurring 3, therefore 26341 is not divisible by 3

8 0
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