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Fofino [41]
2 years ago
14

A bag contains 3 red marbles and 6 blue marbles. what is the probability of randomly selecting a red marble

Mathematics
1 answer:
DochEvi [55]2 years ago
4 0
3/9 is your answer. 3+6=9 and there are 3 red marbles.
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What is the degree of the monomial -9x^4?
devlian [24]

Answer:

4 because 4 is the exponent

Step-by-step explanation:

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3 years ago
Thr profits in December were 5 times the profits in June. If Decembers profit were $3455 how much were thw profits in June?
dedylja [7]
Your Question:
Thr profits in December were 5 times the profits in June. If Decembers profit were $3455 how much were thw profits in June? Our Teams Answer: $3,455 divided by 5 so your answer would be $691

Our Teams aim's to please and We are happy to help! Its the answer is wrong please notify us quickly for we can fix it quickly! Thanks :)

-ExperimentsDIY
7 0
3 years ago
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Need some help. :/
Allushta [10]
I don’t have any graph paper but on 2x draw a line from the very right to the very left and shade the entire left side. And then on -3 draw a line from the very top to the very bottom and shade the top part. Unless you don’t shade it then don’t do that but that’s what you have to do
4 0
2 years ago
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>

7 0
2 years ago
If y = 5x + 9 were changed to y = 5x + 6, how would the graph of the new function compare with the first one?
Talja [164]
The inital value would be different. in 5x+9 it would cross the yaxis at 9 and the other at 6
5 0
3 years ago
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