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3 years ago

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For which equations is x = 9 a possible solution? Check all that apply. |x| = 9 –|x| = 9 –|–x| = 9 –|–x| = –9 |x| = –9 |–x| = 9

|–x| = –9

1 answer:

saul85 [17]3 years ago

4 0

Answer:

918273727

kdiwow

8181...

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In a certain state, car license plates are numbered in the form of LL NNNN, where L stands for a letter of the alphabet and N st

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L=26
N=10
So it's 26*26*10*10*10*10=6760000

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3 years ago

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solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. x-2y+

Alekssandra [29.7K]

Answer:

x = -5, y = -6, z = -3

Step-by-step explanation:

Given the system of three equations:

$\left\{\begin{array}{l}x-2y+3z=-2\\6x+2y+2z=-48\\x+4y+3z=-38\end{array}\right.$

Write the augmented matrix for the system of equations

$\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)$

Find the reduced row-echelon form of the augmented matrix for the system of equations:

$\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)\sim \left(\begin{array}{ccccc}1&-2&3&|&-2\\0&-14&16&|&36\\0&-6&0&|&36\end{array}\right)\sim \left(\begin{array}{ccccc}1&3&-2&|&-2\\0&16&-14&|&36\\0&0&-6&|&36\end{array}\right)$

Thus, the system of three equations is

$\left\{\begin{array}{r}x+3z-2y=-2\\16z-14y=36\\-6y=36\end{array}\right.$

From the last equation:

$y=-6$

Substitute it into the second equation:

$16z-14\cdot (-6)=36\\ \\16z=36-84\\ \\16z=-48\\ \\z=-3$

Substitute y = -6 and z = -3 into the first equation:

$x+3\cdot (-3)-2\cdot (-6)=-2\\ \\x=-2+9-12\\ \\x=-5$

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3 years ago

The two triangles are similar.

Neko [114]

Given that the triangles are similar.

When triangles are similar, all three pairs of corresponding sides are in the same ratio.

Similarly in the given triangles, the ratio of the corresponding sides are equal.

$\frac{8}{2x-2}= \frac{8+6}{3x}$

$\frac{8}{2x-2}= \frac{14}{3x}$

Cross multiplying, we get

$8 \times 3x = 14 \times (2x-2)$

$24x = 28x-28$

$-4x = -28$

Dividing both sides of the equation by -4.

So, we get

x = 7.

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2 years ago

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a laptop's screen is 8 inches long. if the diagonal measures 10 inches, what is the width of the laptop

Ray Of Light [21]

The width of the laptop is 6 inches long.

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2 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

<u><em>a) The probability that exactly 4 flights are on time is equal to 0.0313</em></u>

<u><em></em></u>

<u><em>b) The probability that at most 3 flights are on time is equal to 0.0293</em></u>

<u><em></em></u>

<u><em>c) The probability that at least 8 flights are on time is equal to 0.00586</em></u>

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

<em>United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures. </em>

<em>a) The probability that exactly 4 flights are on time is = </em>

<em>b) The probability that at most 3 flights are on time is = </em>

<em>c)The probability that at least 8 flights are on time is =</em>

<h2>Solution to the problem</h2>

<u><em>a) Probability that exactly 4 flights are on time</em></u>

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

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

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

<em><u>b) Probability that at most 3 flights are on time</u></em>

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

<em><u>c) Probability that at least 8 flights are on time </u></em>

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

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3 years ago