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inn [45]
2 years ago
8

Please help me. justify your answers!​

Mathematics
2 answers:
azamat2 years ago
7 0

Answer:

g. True

h. False

i. False

Step-by-step explanation:

<em>g.</em>

\frac{1}{14} is half of \frac{1}{7}

This would be true because half of a number can be found by multiplying by \frac{1}{2}

If you multiplied \frac{1}{7} * \frac{1}{2}

The numerator is 1 * 1 which is 1

And then the denominator is 7 * 2 which is 14

That would give you \frac{1}{14}

<em>h.</em>

The products of two reciprocals is 0

This is not true

Reason being, reciprocals are just the number flipped. If you times the number by itself, you don't get 0, you get 1.

Ex: \frac{2}{8}

Its reciprocal is \frac{8}{2}

So multiply them together

\frac{2}{8}* \frac{8}{2}

2 * 8 = 16

8 * 2 = 16

\frac{16}{16} = 1

<em>i.</em>

A whole number does not have a reciprocal

This is incorrect. If you think about it, a whole number is just the number, over 1

Ex: \frac{6}{1}

This is still a whole number even though its written in fraction form because if you divide 6 by 1, your still going to get 6.

If you an write it as a fraction, it has a reciprocal.

The reciprocal in this case would be \frac{1}{6}

I hope this helps, don't be afraid to reach out with any further questions!

laila [671]2 years ago
4 0

Answer:

T, F, F

Step-by-step explanation:

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For a certain population of men, 8 percent carry a certain genetic trait. For a certain population of women, 0.5 percent carry t
slavikrds [6]

Answer:

C) 150 men and 100 women

D) 200 men and 2000 women

E) 1000 men and 1000 women

Step-by-step explanation:

Hello!

To compare the proportion of people that carry certain genetic trait in men and woman from a certain population two variables of study where determined:

X₁: Number of men that carry the genetic trait.

X₁~Bi(n₁;p₁)

X₂: Number of women that carry the genetic trait.

X₂~Bi(n₂;p₂)

The parameter of interest is the difference between the population proportion of men that carry the genetic trait and the population proportion of women that carry the genetic trait, symbolically: p₁-p₂

To be able to study the difference between the population proportions you have to apply the Central Limit Theorem to approximate the distribution of both sample proportions to normal.

<u><em>Reminder:</em></u>

Be a variable with binomial distribution X~Bi(n;p), if a sample of size n is taken from the population in the study. Then the distribution of the sample proportion tends to the normal distribution with mean p and variance (pq)/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

So for both populations in the study, the sample sizes should be

n₁ ≥ 30

n₂ ≥ 30

Also:

Both samples should be independent and include at least 10 successes and 10 failures.

Both populations should be at least 20 times bigger than the samples. (This last condition is to be assumed because without prior information about the populations is impossible to verify)

  • If everything checks out then (p'₁-p'₂)≈N(p₁-p₂; p(1/n₁+1/n₂))

<u>The options are:</u>

A) 30 men and 30 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 30*0.08= 2.4

Failures: y₁= n₁*q₁= 30*0.92= 27.6

Population 2

Successes: x₂= n₂*p₂= 30*0.5= 15

Failures: y₂= n₂*q₂= 30*0.5= 15

The second condition is not met.

B) 125 men and 20 women

n₁ ≥ 30 but n₂ < 30

Both samples are independent but n₂ is not big enough for the approximation.

C) 150 men and 100 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 150*0.08= 12

Failures: y₁= n₁*q₁= 150*0.92= 138

Population 2

Successes: x₂= n₂*p₂= 100*0.5= 50

Failures: y₂= n₂*q₂= 100*0.5= 50

All conditions are met, an approximation to normal is valid.

D) 200 men and 2000 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 200*0.08= 16

Failures: y₁= n₁*q₁= 200*0.92= 184

Population 2

Successes: x₂= n₂*p₂= 2000*0.5= 1000

Failures: y₂= n₂*q₂= 2000*0.5= 1000

All conditions are met, an approximation to normal is valid.

E) 1000 men and 1000 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 1000*0.08= 80

Failures: y₁= n₁*q₁= 1000*0.92= 920

Population 2

Successes: x₂= n₂*p₂= 1000*0.5= 500

Failures: y₂= n₂*q₂= 1000*0.5= 500

All conditions are met, an approximation to normal is valid.

I hope this helps!

6 0
3 years ago
What are the solutions of 2x^4-5x^3+60x^2+180x-432
swat32
The polynomial can't be factored because they are all rational numbers.
5 0
3 years ago
2852.1 rounded to the nearest tenth of a mile
katrin2010 [14]
2852.1 rounded to the nearest tenth of a mile is still 2852.1
5 0
3 years ago
Read 2 more answers
A 0.40 kg bird is flying at a constant speed of 8.0 m/s. What is the bird's kinetic energy?
nalin [4]

Answer:

The correct answer is 12.8 J.

Step-by-step explanation:

To solve this problem, we must remember how to calculate the kinetic energy of an object.

The kinetic energy is represented by the formula K = 1/2 * m * v^2, where m represents the mass of the object and v represents the speed or velocity of the object.  If we plug in the given values into the formula, we get:

K = 1/2 * m * v^2

K = (1/2) * (0.40) * (8.0)^2

Our first step is to square the velocity.  After doing this step, we get the following:

K = (1/2) * (0.40) * (64)

Finally, we can perform the multiplication, to get:

K = 12.8

The unit for kinetic energy is joules, so the correct answer is 12.8 J.

Hope this helps!

8 0
3 years ago
Read 2 more answers
Use differentials to find an approximate value for <img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B65%7D%20" id="TexFormul
Sveta_85 [38]
: Let y = f(x) =  x^1/3  
Then dy = 1/3*x^(−2/3) dx 
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We take x = 64 and dx = ∆x = 1  
This gives dy = 1/3*(64)^(−2/3)* (1) =  1/48  
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6 0
3 years ago
Read 2 more answers
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