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3 years ago

9

A local post office sells stamps in packs of 4,6,and 7. Andy bought se era packs of 6. Harry bought several packs of 7. Erin bou

gh several packs of 4. Each of the friends ended up with same amount of stamps. What is the smallest number of stamps th at each person could have purchased?

2 answers:

Whitepunk [10]3 years ago

8 0

Andy bought 21 packs. Harry bought 12 packs. Erin bought 21 packs.

Scilla [17]3 years ago

7 0

<span>Andy bought 21 packs. Harry bought 12 packs. Erin bought 21 packs and that is the answer for all of them. :)</span>

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A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11. Use the TI-84 Plus calculator. (a) I

sukhopar [10]

Answer:

a) We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

b) There is a 15.97% probability that x will be between 54 and 55.

c) The 47th percentile of x is $X = 52.877$ .

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11.

This means that $\mu = 53$ .

We have to find the standard deviation of the sample, that is:

$\sigma = \frac{11}{\sqrt{45}} = 1.64$

(a) Is it appropriate to use the normal distribution to find probabilities for x?

We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

(b) Find the probability that x will be between 54 and 55.

This is the pvalue of the Z score when X = 55 subtracted by the pvalue of the Z score when X = 54.

X = 55

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 53}{1.64}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.88877.

X = 54

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{54 - 53}{1.64}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.72907.

So, there is a 0.88877 - 0.72907 = 0.1597 = 15.97% probability that x will be between 54 and 55.

(c) Find the 47th percentile of x. Round the answer to at least two decimal places.

This is the value of X when Z has a pvalue of 0.47;

This is between $Z = -0.07$ and $Z = -0.08$ . So we use $Z = -0.075$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.075 = \frac{X - 53}{1.64}$

$X = 52.877$

The 47th percentile of x is $X = 52.877$ .

8 0

3 years ago

What is likely to be the subject of a direct geometric proof

notka56 [123]

I think that it is geometry.

4 0

3 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

In a school election, Juan received 4 times as many votes as Wayne, Neal recurved twenty less votes than Juan, and Kerry got hal

77julia77 [94]

Votes received by Wayne is 112

<em><u>Solution:</u></em>

To find: votes received by Wayne

Let the vote received by Wayne be "x"

<em><u>Juan received 4 times as many votes as Wayne</u></em>

Therefore,

Juan votes = 4 times as many votes as Wayne

Juan votes = 4x ---- eqn 1

<em><u>Neal received twenty less votes than Juan</u></em>

Neal votes = twenty less votes than Juan

Neal votes = Juan votes - 20

Neal votes = 4x - 20 ---- eqn 2

<em><u>Kerry got half as many votes as Neal</u></em>

Kerry votes = half of neal votes

Kerry votes = $\frac{4x - 20}{2}$ ---- eqn 3

The total votes cast in the election was 1,202

Wayne votes + Juan votes + Neal votes + Kerry votes = 1202

Plug in eqn 1 , eqn 2, eqn 3

$x + 4x + 4x - 20 + \frac{(4x - 20)}{2} = 1202\\\\2x + 8x + 8x - 40 + 4x - 20 = 1202 \times 2\\\\22x - 60 = 2404\\\\22x = 2404 + 60\\\\22x = 2464\\\\x = 112$

Therefore votes received by Wayne is 112

4 0

3 years ago

If (a+10) and (a+20)are supplementary angles,find them

motikmotik

Given:

(a+10) and (a+20) are supplementary

To find:

Each angle

Steps:

if 2 angles are supplementary that means they add up to 180°

(a + 10) + (a + 20) = 180

a + 10 + a + 20 = 180

2a + 30 = 180

2a = 180 - 30

2a = 150

a = 150/2

a = 75°

Now let's find value of each angle,

a + 10 = 75 + 10

= 85°

a + 20 = 75 + 20

= 95°

Therefore, the value of each angle is 85° and 95° respectively

Happy to help :)

If you need any help feel free to ask

4 0

3 years ago