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pogonyaev
3 years ago
8

(05.08A)Triangle ABC is transformed to similar triangle A′B′C′ below:

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:

1st option is correct i.e., 1 over 2

Step-by-step explanation:

Given that Triangle ABC transformed to Triangle A'B'C'.

Coordinated of Triangle ABC are A(2,6) , B(2,4) and C(4,4)

Coordinated of Triangle A'B'C' are A'(1,3), B'(1,2) and C'(2,2)

Scale factor of Dilation is the no. of times coordinates of vertices of a figure increase or decrease.i.e.,

if A(a,b) ⇒ A'(c,d)

where, a.k=c & b.k=d then k is scale factor of dilation.

Value of k is same for all vertices.

For coordinates of A & A' ,we have 2.k=1\implies k=\frac{1}{2} \:\: and \:\: 6.k=3\implies k=\frac{1}{2}

For coordinates of B & B' ,we have 2.k=1\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}

For coordinates of C & C' ,we have 4.k=2\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}

Therefore, the scale factor of dilation is \:\frac{1}{2}\:  since in all cases k=\frac{1}{2}

1st option is correct i.e., 1 over 2

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Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 30, p = 0.626

So

\mu = E(X) = np = 30*0.626 = 18.78

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is P(X \leq 16 + 0.5) = P(X \leq 16.5), which is the pvalue of Z when X = 16.5. So

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

Z = \frac{16.5 - 18.78}{2.65}

Z = -0.86

Z = -0.86 has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

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3 years ago
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Answer:

(-2, 0)

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the y-intercept is (-2, 0), therefore that the solution.

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