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dsp73
3 years ago
6

26. A positive whole number is called stable if at least one of its digits has the same value

Mathematics
1 answer:
Kisachek [45]3 years ago
5 0

Answer:

<h2>One</h2>

Step-by-step explanation:

Given the value 78247 a s a stable number because at least one of its digits has the same value  as its position in the number. The 4th number in the value is 4, this makes the number a stable number.

The following are the 3-digits stable numbers that appears in 78247

The first number is 824. This digits are stable numbers because 2 as a number is situated in the same place as the number (2nd position).

Hence, there are only 1 stable 3-digit numbers in the value 78247 since only a value exists as 2 in the value and there is no 1 and 3 in the value.

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A prospective mother and a prospective father are hemophiliacs. (Enter your answers as fractions.)
Katena32 [7]

So remember that Hemophilia is <u>a recessive, x linked trait.</u>

For a woman to have hemophilia, she must have the trait linked to both x chromosomes. For a man to have hemophilia, he must have the trait linked to his singular x chromosome.

For this, I will be making a Punnett Square to determine the Possibilities:

\left[\begin{array}{cccc}&&X_h&Y\\&&-&-\\X_h&|&X_hX_h&X_hY\\X_h&|&X_hX_h&X_hY\end{array}\right]

<h3>A.</h3>

Since there is a 1/2 chance for their offspring to be a boy, and of that 1/2 both would be hemophiliacs, <u>there is a 1/2 chance for the child to be a hemophiliac male.</u>

<h3>B.</h3>

As previously mentioned, for a woman to have hemophilia, they would need to have that trait attached to both x chromosomes. Since there is a 1/2 chance for their offspring to be a girl, and of that 1/2 both would be hemophiliacs (since they both carry the trait on both chromosomes), <u>there is a 1/2 chance that they will have a hemophiliac female.</u>

<h3>C.</h3>

<u>So a carrier is someone who carries a recessive trait, but it isn't displayed due to the dominant trait masking it. With x-linked traits, only women can be carriers since they carry more than one x chromosome.</u> What this asks is the probability of an offspring having the trait attached to only 1 of the x chromosomes. Looking at the girls, since both carry the traits on both x chromosomes, <u>there is 0 chance of a carrier.</u>

<h3>D.</h3>

So symptom free is as it seems, without the hemophilia trait. Looking at the table, since all the offspring contain the hemophilia trait, <u>there is 0 chance that any of their offspring will go symptom free.</u>

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Suppose that the value of a stock varies each day from $13 to $24 with a uniform distribution. (a) Find the probability that the
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Answer:

a. P= 0.6364

b. P = 0.3636

c. Q = $21.25

d. P = 0.5

Step-by-step explanation:

given data

value of a stock varies = $13 to $24

solution

P (stock value is more than $17)

P = \frac{(24-17)}{(24-13)}

P =  \frac{7}{11}

P = 0.6364

and

P (value of the stock is between $17 and $21)

P = \frac{(21-17)}{(24-13)}

P = \frac{4}{11}

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and  

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\frac{(24 - Q)}{11} = 0.25

(24 - Q) = 2.75

so

Q = $21.25

and

P(X > 20 | X > 16)

P = \frac{(24-20)}{(24-16)}

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P = 0.5

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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