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MaRussiya [10]
3 years ago
12

The average smoker typically pays 15% higher life insurance costs than the average non-smoker. If the monthly cost to a non-smok

er is $500 dollars per month, how much would the average smoker pay per month for the same life insurance policy?
a) $500.15
b) $515.00
c) $533.00
d) $575.00
Mathematics
1 answer:
Licemer1 [7]3 years ago
5 0
C the is the correct answer
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Solve The Equation <br> 4x×9y=7<br> 4x-9y=9
hoa [83]

Answer:

\large\boxed{x=\dfrac{9}{8}-\dfrac{\sqrt{109}}{8},\ y=-\dfrac{1}{2}-\dfrac{\sqrt{109}}{18}}\\or\\\boxed{x=\dfrac{9}{8}+\dfrac{\sqrt{109}}{2},\ y=-\dfrac{1}{2}+\dfrac{\sqrt{109}}{18}}

Step-by-step explanation:

\left\{\begin{array}{ccc}4x\times9y=7&(1)\\4x-9y=9&(2)\end{array}\right\\\\(2)\\4x-9y=9\qquad\text{subtract}\ 4x\ \text{from both sides}\\-9y=-4x+9\qquad\text{change the signs}\\9y=4x-9\qquad\text{substitute it to (1)}\\\\4x(4x-9)=7\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\(4x)(4x)+(4x)(-9)=7\\(4x)^2-36x=7\\(4x)^2-2(4x)(4.5)=7\qquad\text{add}\ 4.5^2\ \text{to both sides}\\(4x)^2-2(4x)(4.5)+4.5^2=7+4.5^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2

(4x-4.5)^2=7+20.25\\(4x-4.5)=27.25\to 4x-4.5=\pm\sqrt{27.25}\\\\4x-\dfrac{45}{10}=\pm\sqrt{\dfrac{2725}{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{2725}}{\sqrt{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25\cdot109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{5\sqrt{109}}{10}\qquad\text{add}\ \dfrac{45}{10}\ \text{to both sides}\\\\4x=\dfrac{45}{10}\pm\dfrac{5\sqrt{109}}{10}

4x=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 4}\\\\x=\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\\\\\text{Put the values of}\ x\ \text{to (2):}\\\\9y=4\left(\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\right)-9\\\\9y=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}-\dfrac{18}{2}\\\\9y=-\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 9}\\\\y=-\dfrac{1}{2}\pm\dfrac{\sqrt{109}}{18}

8 0
3 years ago
Will 4 5/10 x 6/17 be larger, smaller, the same or impossible to tell than 4 5/10?
Delicious77 [7]
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⁹/₂ × ⁶/₁₇
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The product of 4¹/₂ and ⁶/₁₇ will be smaller than 4¹/₂.
3 0
3 years ago
Derive this with respect to x<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B%281%20%2B%20%20secx%29%20%7D%20" id="Tex
Elenna [48]

~~~~\dfrac{d}{dx} \left(\dfrac{3}{1 + \sec x} \right)\\\\\\=3 \dfrac{d}{dx} \left( \dfrac 1{ 1+ \sec x} \right)\\\\\\=3 \dfrac{d}{dx} (1+ \sec x)^{-1}\\\\\\=3 (-1) (1 + \sec x )^{-1 -1} \dfrac{d}{dx}( 1 + \sec x)\\\\\\=-3(1 + \sec x)^{-2}  ( 0 + \sec x  \tan x)\\\\\\=-\dfrac{3\sec x \tan x}{(1 + \sec x)^2}

8 0
1 year ago
Seven different names were put into a hat. A name chosen 100 times. And the name Michael is chosen 9 times. What is experimental
iris [78.8K]

Answer:

Experimental probability = 9 / 100

Theoretical probability = 1/7

Step-by-step explanation:

The experimental probability of an event is written as :

P = number of times event occur / total number of trials

Experimental probability is based on the outcome of an experiment that has taken place.

Theoretical probability is based on expected outcome of an event.

P = number of expected or required outcomes / total possible outcomes

Experimental probability of choosing the name Michael is :

Number of times Michael is chosen / number of trials

Experimental probability = 9 / 100 = 0.09

Theoretical probability of choosing michael:

Number of names in hat = 7

Number of names called Michael in hat = required outcome = 1

P = 1 / 7

If the number of names in hat was different from 7, then the theoretical probability will change.

Also, if the number of times a name was chosen was different from 100, then number of trials will also change and hence, the experimental probability.

4 0
2 years ago
The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh
mr_godi [17]

Answer:

a) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

b) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number  of defects expected in a production process. Assume a production process produces  items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected  number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus  one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces  will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10,0.15)  

Where \mu=10 and \sigma=0.15

We can calculate the probability of being defective like this:

P(X

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And if we replace we got:

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or  greater than 10.15 ounces being classified as defects.

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

Part c What is the advantage of reducing process variation, thereby causing process control  limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0
3 years ago
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