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andrezito [222]
3 years ago
14

Solve the combined inequality

Mathematics
2 answers:
natta225 [31]3 years ago
6 0

Answer:

6.72<x<10.08

Step-by-step explanation:

multiply 1.12 and 6 for the low end

multiply 1.12 and 9 for the high end

Inessa05 [86]3 years ago
3 0

Answer:

6.72<=X<=10.08

Step-by-step explanation:

6*1.12<=X<=9*1.12

We put 6*1.12 because that is the least she would pay, putting it into the lower end.

We put 9*1.12 because that is the most she would pay, putting it into the higher end.

6.72<=X<=10.08

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Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random
yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we have that the mean is \mu*n and the standard deviation is s = \sigma \sqrt{n}

In this problem, we have that:

\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{4000 - 3500}{160}

Z = 3.13

Z = 3.13 has a pvalue of 0.9991

X = 3000

Z = \frac{X - \mu}{s}

Z = \frac{3000 - 3500}{160}

Z = -3.13

Z = -3.13 has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

5 0
3 years ago
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\&#10;P(x)=3x^3-5x^2-14x-4\\\\&#10;D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\&#10;\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\&#10;3x^3-5x^2-14x-4=0\\\\&#10;

\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
3 years ago
What is the domain and range of the relation shown in the table?
Leokris [45]

Answer:

domain: {-12, -8, 0, 1}   range: {0, 8, 12}

Step-by-step explanation:

domain are of all the input values shown on the x-axis. The range is the set of possible output  shown on the y-axis.

5 0
3 years ago
Make a number line and mark the points that represent the following values of x. 10∙x=0
djyliett [7]

The point on a number line is the location of the point.

<h3>How to draw the number line</h3>

The points are given as:

x = 10 and x = 0

This means that we mark a point at the point 10 on the number line, and another point at the point 0 on the same number line

However, the point 0 would come before point 10 because 0 is less than 10

See attachment for the number line

Read more about number lines at:

brainly.com/question/10851163

6 0
2 years ago
Consuela and her three friends order 2 pizzas. Consuela cuts each pizza into 8 equal slices. She saves 2 slices of pizza for her
Akimi4 [234]
Let
x-------------->  the number of slices eating by Consuela  and her <span>three friends

we know that
[total slices]=8*2=16 slices
</span><span>[slices of pizza for her older brother]=2
x=16-2=14 slides
</span><span>Consuela and her friends share the rest of the pizza, each eating the same number of slices---------> 14/4= 3.5 slides of pizza each  (</span><span>this part is not necessary, but it is shown in a didactic way)
</span><span>
the question is 
</span><span>What fraction of the 2 pizzas is left over?
</span>[slices of pizza for her older brother]/[total slices]=2/16=1/8
<span>
the answer is (1/8)

</span>
8 0
3 years ago
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