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Neporo4naja [7]
3 years ago
6

Write an equation for: melissa has 6 times as many quarters as Michelle. Together they have 896 quarters. How many quarters, q,

does Michelle have? Please and thanks.
Mathematics
1 answer:
butalik [34]3 years ago
6 0
Okay, so in this equation we will use the following variables:

a: for the number of quarters Michelle has
b: for the number of quarters Melissa has
q: for total number of quarters

So first, we will write out the statement "Melissa has 6 times as many quarters as Michelle" in the form of an equation:

6 times the number of quarters Michelle has is equal to the number of quarters Melissa has, or:
6(a)=b

So, we now know that Melissa has  6 times as many quarters as Michelle. 

So, if the have 896 quarters total, then you can write the problem as the number of quarters Michelle has added to the number of quarters Melissa has. This can be expressed as:

a+b=q

Since we know that Melissa has 6 times the quarters as Michelle, 6a is equivalent to b:

(6×a)=b

Since we know the total number of quarters is 896, we can rewrite it as this:

a+(6×a)= 896

Now that we know this, we can solve the problem with the following steps:

6a+a=7a

7a=896

896÷7= 128

a= 128

So, now that we know that know that Michelle has 128 quarters, we can find out how many Melissa has:

128×6= 768

Melissa has 768 quarters.

Now we check our work:

128+ (6×128)= 768

If everything is correct the following should be true:

a=128

and 

b=768

So, Michelle has 128 quarters.












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(a) Approximate the percent error in computing the area of the circle: 4.5%

(b) Estimate the maximum allowable percent error in measuring the circumference if the error in computing the area cannot exceed 3%: 0.6 cm

Step-by-step explanation:

(a)

First we need to calculate the radius from the circumference:

c=2\pi r\\r=\frac{c}{2\pi } \\c=8.9 cm

I leave only one decimal as we need to keep significative figures

Now we proceed to calculate the error for the radius:

\Delta r=\frac{dt}{dc} \Delta c\\\\\frac{dt}{dc} =    \frac{1}{2 \pi } \\\\\Delta r=\frac{1}{2 \pi } (1.2)\\\\\Delta r= 0.2 cm

r = 8.9 \pm 0.2 cm

Again only one decimal because the significative figures

Now that we have the radius, we can calculate the area and the error:

A=\pi r^{2}\\A=249 cm^{2}

Then we calculate the error:

\Delta A= (\frac{dA}{dr} ) \Delta r\\\\\Delta A= 2\pi r \Delta r\\\\\Delta A= 11.2 cm^{2}

A=249 \pm 11.2 cm^{2}

Now we proceed to calculate the percent error:

\%e =\frac{\Delta A}{A} *100\\\\\%e =\frac{11.2}{249} *100\\\\\%e =4.5\%

(b)

With the previous values and equations, now we set our error in 3%, so we just go back changing the values:

\%e =\frac{\Delta A}{A} *100\\\\3\%=\frac{\Delta A}{249} *100\\\\\Delta A =7.5 cm^{2}

Now we calculate the error for the radius:

\Delta r= \frac{\Delta A}{2 \pi r}\\\\\Delta r= \frac{7.5}{2 \pi 8.9}\\\\\Delta r= 0.1 cm

Now we proceed with the error for the circumference:

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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
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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!

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