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4 years ago

Write the ordered pair that represents yz. Then find the magnitude of yz Y(3,1), Z(0,4)

Mathematics

1 answer:

NikAS [45]4 years ago

5 0

Answer:

(-3, 3), $2\sqrt{2}$

Step-by-step explanation:

I'm assuming this is a vector problem. See the graph for the vector from Y to Z (the tail is at point Y, the head is at point Z).

This vector is equivalent to one which is the same length and in the same direction with its tail at the origin. That vector's head is at (-3, 3) -- the blue one.

To get that, subtract the tail of the original vector from the head of the original vector.

(0, 4) - (3, 1) = (0 - 3, 4 - 1) = (-3, 3)

The magnitude of the blue vector is its length. Use the Distance Formula.

$\sqrt{(x_2 - x_1)^2+(y_2-y_1)^2} \\ \sqrt{(-3-0)^2+(3-0)^2} \\ \sqrt{9+9} \\ \sqrt{18}$

This simplifies to $3\sqrt{2}$

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What is the probability that a random sample of 1414 second grade students from the city results in a mean reading rate of more

Sonbull [250]

Answer with explanation:

Number of second grade students who are considered for sample =1414

Number of words that is being read in a Minute = 9292 words per minute

If number of students will be more than half of 1414, that is if, they can read or their reading rate is more than 9292 words per minute, then we can say that the sample chosen is Appropriate.

⇒Required Probability,that sample of 1414 students can read 9292 words per minute

$=\frac{707+x}{1414} \text{Where }, [1 \leq x\leq 707]$

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3 years ago

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Sonja [21]

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3 years ago

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X < –3

slavikrds [6]

It’s the last one because it’s bigger number and sign

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3 years ago

The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1000 lbs

ElenaW [278]

Answer:

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 1000, \sigma = 200$