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ss7ja [257]
3 years ago
15

Which statements are true about the ordered pair (2, 3) and the system of equations?

Mathematics
1 answer:
mel-nik [20]3 years ago
8 0

substitute the ordered pair (2, 3 )  into the 2 equations

(3 × 2 ) + (4 × 3 ) = 6 + 12 = 18 → True

(2 × 2 ) - (2 × 3 ) = 4 - 6 = - 2 → False

Hence the ordered pair (2, 3 ) is not a solution to the system of linear equations.

When (2, 3 ) is substituted into the first equation, the equation is true.

When (2, 3 ) is substituted into the second equation, the equation is false.



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A car takes 4 hours to cover a certain distance, travelling at a speed of 120 km/h. How long will the car take if it travels at
muminat

Answer:

3/8 hours

Step-by-step explanation:

distance=speed/time=120/4=30km

time=distance/speed=30/80=3/8

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Solve for the variable x<br><br> m = -12g + x
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Answer:

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I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an
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Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

\mu = 0 - 5 = -5

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

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

Z = \frac{1 - (-5)}{\sqrt{13}}

Z = \frac{6}{\sqrt{13}}

Z = 1.66

Z = 1.66 has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

5 0
3 years ago
What is the measure of ∠B?
DochEvi [55]
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