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disa [49]
2 years ago
4

A line segment has endpoints (-4,-6) and (-6,4). Which reflection will produce an image with end points at (4,-6) and (6,4)?

Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
8 0
Reflection about x axis: (x, y) ⇒ (x, -y)
Reflection about y axis: (x, y) ⇒ (-x, y)
Reflection about line y = x: (x, y) ⇒ (y, x)
Reflection about line y = -x: (x, y) ⇒ (-y, -x)
Reflection about origin: (x, y) ⇒ (-x, -y)

(-4, -6), (-6, 4) ⇒ (4, -6), (6, 4)
This is a reflection about the y axis because they follow (-x, y).
The x coordinate becomes a negative (in both cases they become positive because two negatives) and the y coordinate stays the same.
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Answer:

The answer is D

Step-by-step explanation:

When you divide 110 by 8 that equals 13.75 but when you divide 120 by 8 you get 15.

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Angle ABD measures (4x + 10)o. Angle ACD measures (5x − 2)o.
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96°

Step-by-step explanation:

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The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm
sweet-ann [11.9K]

Answer:

a) P(12.99 ≤ X ≤ 13.01) = 0.3840

b) P(X ≥ 13.01) = 0.3075

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the cental limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 13, \sigma = 0.08

(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.

Here we have n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02

This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.

X = 13.01

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

By the Central Limit Theorem

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

Z = \frac{13.01 - 13}{0.02}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

X = 12.99

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

Z = \frac{12.99 - 13}{0.02}

Z = -0.5

Z = -0.5 has a pvalue of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?

P(X ≥ 13.01) =

This is 1 subtracted by the pvalue of Z when X = 13.01. So

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

Z = \frac{13.01 - 13}{0.02}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

1 - 0.6915 = 0.3075

P(X ≥ 13.01) = 0.3075

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A cylindrical rainwater tank is 1.5 m tall with a diameter of 1.4 m. What is the maximum volume of rainwater it can hold??
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To work out the volume of a prism, you multiply the area of cross-section by the height (or length).

So for a cylinder, you work out the area of the circle and then multiply it by the height.

Area of circle (or cross-section) = π × radius²

                                                = π × 0.7²

                                                = 0.49π   m²

Now to get the volume of the cylinder, you times this area of the cross-section by the height of the cylinder:

Volume = 0.49π × 1.5

            = 2.3 m³  (accurate to 2 decimal places)

---------------------------------------------------------------

Answer:

The maximum volume of rainwater the cylinder can hold is:

<u> </u><u>2.3 </u><u>m³</u>

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