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harkovskaia [24]
2 years ago
10

Consider the points in the table below. Determine whether each point would be less than, greater than, or exactly equal to 15 un

its from the origin if plotted in the coordinate plane.

Biology
1 answer:
mart [117]2 years ago
4 0

The categories of the points are

  • Greater than 15 units from the origin: B(-5,15) and E(-8.-14)
  • Exactly 15 units from the origin: C(12,-9)
  • Less than 15 units from the origin: A(11,-6) and D(10.4)

<h3>How to categorize the points?</h3>

The distance of a point from the origin is calculated using:

d = \sqrt{x^2 + y^2}

Using the above formula, and the given coordinates of points A to E, we have:

A = \sqrt{11^2 + (-6)^2} = 12.53

B = \sqrt{(-5)^2 + (15)^2} = 15.81

C = \sqrt{(12)^2 + (-9)^2} = 15

D = \sqrt{10^2 + 4^2} = 10.77

E = \sqrt{(-8)^2 + (-14)^2} = 16.12

From the above computations, we have:

  • Greater than 15 units from the origin: B(-5,15) and E(-8.-14)
  • Exactly 15 units from the origin: C(12,-9)
  • Less than 15 units from the origin: A(11,-6) and D(10.4)


Read more about distance at:

brainly.com/question/7243416

#SPJ1

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Duchenne muscular dystrophy is an X-linked recessive disease. A phenotypically normal couple wants to start a family. The woman’
Andreyy89

Answer:

\frac{1}{8}

Explanation:

From the question: Duchenne muscular dystrophy is an X-linked recessive disease.

Now for an X-linked recessive disorder to be affected by a male individual; it only requires the presence of only one copy of the recessive allele of the disease to be present BUT in a female individual. both copies of the recessive allele must be present.

Let the X-linked recessive disease(i.e Duchenne muscular dystrophy) be = (ⁿ)

However, we are told that the couples are normal and are unaffected. Therefore;

the male partner (XY) will be:   X^NY

the female partner (XY) will be: X^NX^N

Similarly, the question proceeds by telling us that: the woman's brother has the disease.

Definitely, it's possible that this unaffected woman is a carrier of the disease.

So, if she is a carrier; we have her traits to be: X^NX^n

NOW, if a cross exist between these couples; we have

X^NY     ×        X^NX^n

                     X^N                         Y

X^N                X^NX^N                  X^NY

X^n                 X^NX^n                   X^nY

So, we have offspring as follows:

X^NX^N  = normal unaffected female  

X^NY     = normal unaffected male

X^NX^n   = female carrier for the Duchenne muscular dystrophy disease

X^nY      = affected male with the Duchenne muscular dystrophy disease

So, the probability of the child to be affected with the disease = \frac{1}{4}

Also, the probability that the first child will be a male or a female = \frac{1}{2}

∴

the  probability that the couple’s first child will be affected = \frac{1}{4}*\frac{1}{2}

=\frac{1}{8}

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Answer:

sure,what you need?

Explanation:

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