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valina [46]
3 years ago
13

Please solve with full steps (only question 6)

Mathematics
1 answer:
Stels [109]3 years ago
8 0

Answer:

\large \boxed{\sf \ \ k=1 \ \ }

Step-by-step explanation:

Hello,

First of all, let's notice that even if we do not know the zeros of P(x) we can say that

       (1) \ \alpha + \beta =\dfrac{5}{6} \\ \\ (2) \ \alpha * \beta =\dfrac{k}{6}

<u>But, why !?</u>

As they are the zeros of P(x), we can write:

P(x)=6(x^2-\boxed{\dfrac{5}{6}}x+\boxed{\dfrac{k}{6}})=6(x-\alpha)(x-\beta)=6(x^2-\boxed{(\alpha +\beta)}x+ \boxed{\alpha *\beta} )

And then we can identify the coefficients of the like terms to find the equations (1) and (2).

Now, we have <u>one more equation</u> which is:

       (3) \ \alpha -\beta =\dfrac{1}{6}

(1)+(3) gives:

   \alpha + \beta +\alpha -\beta =\dfrac{5}{6}+\dfrac{1}{6}=\dfrac{6}{6}=1 \\ \\ 2\alpha =1 \ \text{ divide by 2 } \\ \\  \alpha =\dfrac{1}{2}

And we replace in (3) to get the value of the second zero.

   \beta = \dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}

And, finally, from (2), it comes:

   k=6*\alpha *\beta =6*\dfrac{1}{2}*\dfrac{1}{3}=\dfrac{6}{6}=1

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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Answer:

a) P(5.8

b) P(5.8

c) For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.

For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values

Step-by-step explanation:

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".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Let X the random variable that represent the weights of​ full-term newborn babies of a population, and for this case we know the distribution for X is given by:

X \sim N(7,1.2)

a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the meanlong dashthat ​is, between 5.8 and 8.2 ​pounds, or within one standard deviation of the​ mean?

We are interested on this probability

P(5.8

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

P(5.8

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(5.8

b. What is the probability that the average of nine ​babies' weights will be within 1.2 pounds of the​ mean; will be between 5.8 and 8.2 ​pounds?

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(7,\frac{1.2}{\sqrt{9}})

The z score on this case is given by this formula:

z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}

And if we replace the values that we have we got:

z_1=\frac{5.8-7}{\frac{1.2}{\sqrt{9}}}=-3

z_2=\frac{8.2-7}{\frac{1.2}{\sqrt{9}}}=3

For this case we can use a table or excel to find the probability required:

P(5.8

c. Explain the difference between​ (a) and​ (b).

For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.

For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values

3 0
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The length of an object object in feet is equal to 3 times it’s linked in yards the length of wall water side slide is 48 feet r
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<h3><u>Question:</u></h3>

The length of an object in feet is equal to 3 times its length in yards. The length of a waterslide is 48 feet. Write and solve a multiplication equation to find the length of the waterslide in yards.

<h3><u>Answer:</u></h3>

The length of waterslide is 16 yards

<h3><u>Solution:</u></h3>

Given that,

The length of an object object in feet is equal to 3 times it’s length in yards

The length of wall water side slide is 48 feet

Let "y" be the length of object in yards

Write and solve a multiplication equation to find the length of the waterslides in yard

From given,

Length in feet = 3 times length in yards

48 = 3 \times y

<em><u>Therefore, the multiplication equation is:</u></em>

3y = 48

Solve the equation for "y"

Divide both sides of equation by 3

\frac{3y}{3} = \frac{48}{3}\\\\y = 16

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