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3 years ago

Do you know the inequality? Please explain and show answer :)

Mathematics

1 answer:

DerKrebs [107]3 years ago

4 0

If u have a open circle, then the inequality has no equal sign. But if it is a closed circle, the inequality has an equal sign.
Shading to the left means " less then " .
Shading to the right means " greater then " .

(1) u have an open circle on -7 with shading to the left....
x < -7

(2) u have a closed circle on 4.5 with shading to the left....
x < = 4.5 (thats less then or equal)

(3) u have an open circle on -5 with shading to the right...
x > -5

(4) u have an open circle on 1.5 with shading to the right...
x > 1.5

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

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3 years ago

Solving one-step inequalities by multiplying/dividing <img src="https://tex.z-dn.net/?f=14v%20%5Cleqslant%2014" id="TexFormul

liubo4ka [24]

Answer: v <= 1

Step-by-step explanation:

When you have a variable by a number, you do the opposite of multiplying, so you divide

Since there's only 14v you divide 14 from itself leaving just the variable

14 ÷ 14 = 1

You repeat this and divide on the other side

14 ÷ 14 = 1

Since you arent dividing by a negative on both sides, the arrow stays the same

v <= 1

Hope this helps!

5 0

3 years ago

some psychologists believe that a genius should be defined as anyone having an IQ over 140. If IQ scores are normally distribute

denis23 [38]

Answer:

61,239,550

Step-by-step explanation:

We let the random variable X denote the IQ scores. This would imply that X is normal with a mean of 100 and standard deviation of 17. We proceed to determine the probability that an individual chosen at random from the population would be a genius, that is;

Pr( X>140)

The next step is to evaluate the z-score associated with the IQ score of 140 by standardizing the random variable X;

$Pr(X>140)=Pr(Z>\frac{140-100}{17})=Pr(Z>2.3529)$

The area to the right of 2.3529 will be the required probability. This area from the standard normal tables is 0.009314

From a population of 6,575,000,000 the number of geniuses would be;

6,575,000,000*0.009314 = 61,239,550

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3 years ago

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