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Lady bird [3.3K]
4 years ago
9

he seeds in bush bean pods are each the product of an independent fertilization event. Green seed color is dominant to white see

d color in bush beans. If a heterozygous plant with green seeds self-fertilizes, what is the probability that 6 seeds in a single pod of the progeny plant will consist of. 3 green and 3 white seeds
Mathematics
1 answer:
-BARSIC- [3]4 years ago
4 0

Answer:

THE ANSWER IS 0.055

Step-by-step explanation:

The green color seed (P) is dominated the white color seed ( Y).

Given a heterozygous plant with green seed self fertilizes .

The cross will be: PY X PY  = 1 PP: 2 PY: 1YY

Chances of having 3 white seeds out of 6 = 3/3 * 1/3 = = 1/3  .

Chances of having 3 green out of 6 = 1/3 * 1/6 = 1/18 = 0.0555

 

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Simplify 3^8 x 3^4 / 3^2 x 3^8 leaving your answer in index form
Furkat [3]

Answer:

3² or 9

Step-by-step explanation:

Step 1: Add exponents

\frac{3^{12}}{3^{10}}

Step 2: Subtract exponents

You should get 3² because of exponent rules. Remember when you multiply exponents you add the powers and when you divide exponents you subtract the powers.

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3 years ago
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Find x. Round your answer to the nearest tenth of a degree.
solniwko [45]

Answer:

4

Step-by-step explanation:

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4 years ago
It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando
notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

3 0
3 years ago
In the rolling of two fair dice calculate the following: P(Sum of the two dice is 7) = ______
vesna_86 [32]

Answer:

P(Sum of the two dice is 7) = 6/36

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

A fair dice can have any value between 1 and 6 with equal probability. There are two fair dices, so we have the following possible outcomes.

Possible outcomes

(first rolling, second rolling)

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)

(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)

(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)

(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

There are 36 possible outcomes.

Desired outcomes

Sum is 7, so

(1,6), (6,1), (5,2), (2,5), (3,4), (4,3).

There are 6 desired outcomes, that is, the number of outcomes in which the sum of the two dice is 7.

Answer

P(Sum of the two dice is 7) = 6/36

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16.67

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