The sum of ten and the product of 4 and x

Which graph represents the solution set of this inequality?

lesya692 [45]

Answer:

the answer is C

Step-by-step explanation:

4 0

3 years ago

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Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random

Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = $50,000

$\sigma$ = population standard deviation = $25,000

n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P( $\bar X$ > $52,000)

P( $\bar X$ > $52,000) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }$ ) = P(Z > 2.53) = 1 - P(Z $\leq$ 2.53)

= 1 - 0.9943 = 0.0057

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

5 0

3 years ago

The population of Williston is currently 21,200 people. If the population

Pachacha [2.7K]

$21200 + 415(y)$

$27000 = 21200 + 415(y)$

$27000 - 21200 = 21200 - 21200 + 415(y)$

$5800 = 415(y)$

$\frac{5800}{415} = \frac{415(y)}{415}$

$13.9759..... = y$

$it \: will \: take \: about \: 14.0 \: years$

8 0

3 years ago

Fiona wrote the linear equation y = 2/5 X -5. When Henry wrote his equation they discovered that his equation had all the same S

REY [17]

The equation of the graph is given as y = (2/5)x - 5.

You have to figure out which of the choices equals the equation given,
y = (2/5)x - 5.
You could solve each of the answer choices for y, but since each choice is in the format x - (m)y = b, you can put the given equation in that format too.

$y= \frac{2}{5}x - 5 \\ 
5= \frac{2}{5}x-y \\ 
 \frac{2}{5}x - y = 5 \\ 
 (\frac{5}{2}) \frac{2}{5}x - (\frac{5}{2})y = (\frac{5}{2})5 \\ 
x - (\frac{5}{2})y = \frac{25}{2}$

5 0

3 years ago

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Javier simplified the expression below. Find and describe the three mistakes he made.

OverLord2011 [107]

$( \dfrac{20x}{5x^8})^-^3$

Take out the constants

$( \dfrac{20}{5} * \dfrac{x}{x^8} ) ^-^3$

$(4*\dfrac{x}{x^8} ) ^-^3$

Use the rule : $\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}
$

$(4x^1^-^8)^-^3$

$(4x^-^7)^-^3$

Use the negative power rule : ${x}^{-a}=\frac{1}{{x}^{a}}
$

$\frac{1}{4(x^-^7)}^-^3$

$\frac{1}{4^3(x^-^7)}^-^3$

$\dfrac{1}{64x^-^2^1}$

$\dfrac{1}{64* \dfrac{1}{x^2^1}}$

$\dfrac{1}{64 \dfrac{}{x^2^1}}$

$= \dfrac{x^2^1}{64}$

6 0

3 years ago

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