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3 years ago

10

roger sells 23 necklaces at a fair a total of $ 201.25. Each necklaces is sold for the same price. what is the price of one neck

laces ?

1 answer:

MrMuchimi3 years ago

4 0

Answer:

$8.75

Step-by-step explanation:

It is given that,

Roger sells 23 necklaces at a fair a total of $ 201.25.

We need to find the price of 1 necklace. We can use unitary method to find it.

23 necklaces = $201.25.

For cost of 1 necklace, divide $201.25 by 23 as follows :

$x=\dfrac{201.25}{23}\\\\=\$ 8.75$

So, the cost of 1 necklace is $8.75.

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4 years ago

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4 years ago

Find (f/g) (x) for the functions provided: ƒ(x) = x3 − 27, g(x) = 3x − 9

AnnZ [28]

Answer:

$(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)$

Step-by-step explanation:

We have been given that

$f(x)=x^3-27,g(x)=3x-9$

We can use the formula for difference of cubes to simplify the function f(x)

difference of cubes - $a^3-b^3=(a-b)(a^2+ab+b^2)$

$f(x)=x^3-27\\\\=x^3-3^3\\\\=(x-3)(x^2+3x+9)$

And g(x) can be written as

$g(x)=3x-9\\=3(x-3)$

Thus, we have

$(\frac{f}{g})(x)=\frac{(x-3)(x^2+3x+9)}{3(x-3}$

On cancelling the common factors, we get

$(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)$

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3 years ago

The heights of North American women are normally distributed with a mean of 64 inches and a standard deviation of 2inches.a. Wha

Charra [1.4K]

Answer:

a) $P(X>66)=P(Z>1)=1-P(Z$

b) $P(\bar X >66)=P(Z>2)=1-P(Z$

c) $P(\bar X >66)=P(Z>10)=1-P(Z$

Step-by-step explanation:

1) 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".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Part a

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

$X \sim N(64,2)$

Where $\mu=64$ and $\sigma=2$

We are interested on this probability

$P(X>66)$

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:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-\mu}{\sigma})=P(Z>\frac{66-64}{2})=P(Z>1)$

And we can find this probability on this way:

$P(Z>1)=1-P(Z$

3) Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

$P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{4}}}=2)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>2)=1-P(Z$

4) Part c

$P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{100}}}=10)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>10)=1-P(Z$

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The correct answer for this would be 1.00 1.25. Hope this is helpful.

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