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3 years ago

ANSWER FAST WHERE DO THEY GO :Drag the graphs to match each inequality with its solution set.

Mathematics

2 answers:

sweet-ann [11.9K]3 years ago

3 0

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AnnZ [28]3 years ago

3 0

Answer:

First of all, remember that signs $\leq \geq$ are represented as solid circles on the number line, and signs are represented as empty circles on the number line.

Also, $\geq$ and $>$ refers to the solution that is at the right side of the number, and and $\leq$ represents solutions at the left side of the number.

So,

For $p\geq 6$ , the graph is the red line that begins at 6 with solid circle and points to the right side.
For $p>6$ , the graph is the red line that begins at 6 with an empty circle and points to the right side.
For $p$ , the graph is the red line that begins at 6 with an empty circle and points to the left side.
For $p\leq 6$ , the graph is the red line that begins at 6 with solid circle and points to the left side.

You might be interested in

Which point on the x-axis lies on the line that passes

posledela

Answer:

(2,0)

Step-by-step explanation:

For given line AB:

y-intercept = b = -2

slope = m = y₂-y₁/x₂-x₁

= -3/2-(-4) = -3/6 = -1/2

Equation of line AB:

y = (-1/2)x - 2

Finding equation of line that is parallel to line AB and passes through the point C(2,2):

Substituting the slope from line AB into the equation of the line

y = (-1/2)x + b.

Substituting the given point (-2,2) into the x and y values 2 = (-1/2)-2 + b.

Solving for b (the y-intercept) , we get b = 1

Substitute this value for 'b' in the slope intercept form equation y = (-1/2)x + 1.

For x-intercept of the line, we let y = 0

0 = (-1/2)x + 1

x = -1(-2/1)

x = +2

So, the point on the x-axis that lies on the line that passes

through point C and is parallel to line AB is (2,0).

7 0

3 years ago

The hourly median power (in decibels) of received radio signals transmitted between two cities

trasher [3.6K]

Using the lognormal and the binomial distributions, it is found that:

The 90th percentile of this distribution is of 136 dB.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.

There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.

In a lognormal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

In this problem:

The mean is of $\mu = 3.5$ .

The standard deviation is of $\sigma = \sqrt{1.22}$

Question 1:

The 90th percentile is X when Z has a p-value of 0.9, hence X when Z = 1.28.

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

$1.28 = \frac{\ln{X} - 3.5}{\sqrt{1.22}}$

$\ln{X} - 3.5 = 1.28\sqrt{1.22}$

$\ln{X} = 1.28\sqrt{1.22} + 3.5$

$e^{\ln{X}} = e^{1.28\sqrt{1.22} + 3.5}$

$X = 136$

The 90th percentile of this distribution is of 136 dB.

Question 2:

The probability is the p-value of Z when X = 150, hence:

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

$Z = \frac{\ln{150} - 3.5}{\sqrt{1.22}}$

$Z = 1.37$

$Z = 1.37$ has a p-value of 0.9147.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.

Question 3:

10 signals, hence, the binomial distribution is used.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

For this problem, we have that $p = 0.9147, n = 10$ , and we want to find P(X = 6), then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.9147)^{6}.(0.0853)^{4} = 0.0065$

There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.

You can learn more about the binomial distribution at brainly.com/question/24863377

5 0

2 years ago

6,666,666×666,666 /1+2+3+4+5+6+5+4+3+2+1 − 777,777×777,777/1+2+3+4+5+6+7+6+5+4+3+2+1

jeka57 [31]

Answer:

111111 $\times$ 1111110 is the simple expression.

Step-by-step explanation:

The expression to be solved:

$\dfrac{6,666,666\times666,666 }{1+2+3+4+5+6+5+4+3+2+1} - \dfrac{777,777\times777,777}{1+2+3+4+5+6+7+6+5+4+3+2+1}$

First of all, let us solve the first term:

$\dfrac{6,666,666\times666,666 }{1+2+3+4+5+6+5+4+3+2+1}\\\Rightarrow \dfrac{6666666\times666666 }{36}\\\Rightarrow \dfrac{6666666\times666666 }{6\times 6}\\\Rightarrow 1111111\times 111111$

Now, the right term:

$\dfrac{777777\times777777}{1+2+3+4+5+6+7+6+5+4+3+2+1}\\\Rightarrow \dfrac{777777\times777777}{49}\\\Rightarrow \dfrac{777777\times777777}{7 \times 7}\\\Rightarrow 111111 \times 111111$