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3 years ago

Are the triangles similar? How do you know? A. yes, by SAS B. yes, by SSS C. yes, by AA D. no

Mathematics

2 answers:

lord [1]3 years ago

8 0

Answer:

Option D is correct.

No, the triangles are not similar.

Step-by-step explanation:

AA postulates states that two triangles are similar if they have two corresponding angles equal.

Labelled the diagram as shown below:

We know that sum of all the measure of the angles in a triangle is 180 degree.

In triangle ABC:

$\angle A + \angle B + \angle C = 180^{\circ}$

⇒ $30.4^{\circ}+84.6^{\circ}+\angle C = 180^{\circ}$

⇒ $115^{\circ}+\angle C = 180^{\circ}$

Subtract 115 degree from both sides we get;

⇒ $\angle C = 65^{\circ}$

In triangle ABC and PQR

$\angle A = \angle Q = 84.6^{\circ}$

$\angle C \neq \angle R$

i.e $65^{\circ}\neq 66^{\circ}$

⇒These triangles does not satisfy the AA postulates

Therefore, the given triangles are not similar.

WARRIOR [948]3 years ago

4 0

No, because 30.4+84.6+66=181
the sum of a triangle=180
so they are not similar.

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(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

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Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

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The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

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$\sigma$ = standard deviation = 20 minutes

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The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

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