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MAVERICK [17]
3 years ago
5

In the f2 generation of a homozygous round (aa) x homozygous wrinkled (aa) cross in peas, two round seeds are chosen at random.

what is the probability that one is aa and the other aa?
Mathematics
1 answer:
LenaWriter [7]3 years ago
4 0
The answer in this question whose asking for the probability of the two seed are round and the other is wrinkled is 3(1/4)(3/4)2. The probability that two seeds are round (AA) and the other is wrinkled (aa) is 3 (1/4)(3/4) 2.  3 (1/4)(3/4) 2 is the answer in this question. 
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Lexy and Lonnel applied to the same university. They looked up the average SAT score of students admitted to that university. Le
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Answer:

  (a)  ❘-270 - 30❘

Step-by-step explanation:

One score is -270 and the other is +30. The difference is either of ...

  |30 -(-270)|

or

  |-270 -30| . . . . . . matches choice A

4 0
3 years ago
Suppose that 20° of boys opted for mathematics and 40% of girls opted for mathematics. What is the probability that a student op
ankoles [38]

Answer: 30%

Step-by-step explanation:

Let A be the probability of a student opting for mathematics - it consists of either boy opting for mathematics or girl opting for mathematics. As there is "or" we need to sum these probabilities.

P(A) = P(B)* P(M|G) + P(G)*P(M|G)

P(A) = \frac{1}{2} * \frac{20}{100} + \frac{1}{2} * \frac{40}{100}

P(A) = 3/10 = 0.3

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Setler79 [48]

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4 0
3 years ago
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The data shown represent the number of runs made each year during Bill Mazeroski’s career. Check for normality.
const2013 [10]

Answer:

The given data is not normal.

Step-by-step explanation:

We are given the following data:

30, 59, 69, 50, 58, 71, 55, 43, 3,  66, 52, 56, 62, 36, 13, 29, 17, 31

Condition for normality:

Mean = Mode = Median

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{800}{18} = 44.44

Mode is the most frequent observation of the data.

Since all the value appeared once, there is no mode.

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Sorted data: 3, 13, 17, 29, 30, 31, 36, 43, 50, 52, 55, 56, 58, 59, 62, 66, 69, 71

Median =

=\dfrac{9^{th}+10^{th}}{2} = \dfrac{50+52}{2}=51

Since the mean, mode and median of data are not equal, the data is not normal.

7 0
3 years ago
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