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Nostrana [21]
3 years ago
10

Given the preimage and image,find the dilation scale factor.​

Mathematics
1 answer:
Oliga [24]3 years ago
7 0

Answer:

The dilation scale factor is \frac{1}{2}.

Step-by-step explanation:

The image is the dilated form of its preimage if and only if the following conditions are observed:

1) K' = \alpha_{1} \cdot K

2) T' = \alpha_{2} \cdot T

3) P' = \alpha_{3} \cdot P

4) J' = \alpha_{4} \cdot J

5) \alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}

If we know that K = (2, 0), K' = (1, 0), T = (3, 0), T' = (1.5,0), P = (1, 5), P' = (0.5, 2.5), J = (0, 3) and J' = (0, 1.5), then the coefficients are, respectively:

\alpha_{1} = \frac{1}{2}, \alpha_{2} = \frac{1}{2}, \alpha_{3} = \frac{1}{2}, \alpha_{4} = \frac{1}{2}

As \alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}, we conclude that the dilation scale factor applied in the preimage is equal to \frac{1}{2}.

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Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = <em>Number of adults who use their smartphones</em>

The above situation can be represented through the binomial distribution;

P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......

where, n = number of trials (samples) taken = 15 adult smartphones

           r = number of success = at least 4

           p = probability of success which in our question is the % of adults

                 who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X \geq 4)

P(X \geq 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= 1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}

= 1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})

= <u>0.1773</u>

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3 years ago
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7 0
3 years ago
Read 2 more answers
(10-1)(10-2)(10-3)(10-10)
Goryan [66]

Answer:

0

Step-by-step explanation:

(10-1=9)(10-2=8)(10-3=7)(10-10=0)

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6 0
3 years ago
Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grad
AlekseyPX

Answer:

Since the calculated value of F = 1.4397 is less than the critical value of

F (9,9)= 2.4403 we conclude that the  first instructor's variance is smaller and reject H0.

Step-by-step explanation:

1)Formulate the hypothesis that first variance is equal or greater than the second variance

H0: σ₁²≥σ₂²  against the claim  that the first instructor's variance is smaller

 Ha: σ₁²< σ₂²

2) Test Statistic F= s₂²/s₁²

F= 84.8/ 58.9=  1.4397

3)Degrees of Freedom = n1-1= 10-1= 9  and n2 = 10-1= 9

4)Critical value   at 10 % significance level= F(9,9)= 2.4403

5)Since the calculated value of F = 1.4397 is less than the critical value of

F (9,9)= 2.4403 we conclude that the  first instructor's variance is smaller and reject H0.

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