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2 years ago

The rectangle shown is dilated by a scale factor of 4

Mathematics

1 answer:

EastWind [94]2 years ago

5 0

Answer:

60 cm, 60 cm, 32 cm, 32 cm

Step-by-step explanation:

Multiply each length by the scale factor.

15 cm * 4 = 60 cm

8 cm * 4 = 32 cm

The dilated sides AB and CD will have a length of 60 cm.

The dilated sides AC and BD will have a length of 32 cm.

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The blue group did because they ate 7 slices out of 8 while the green team only ate 2 out of 8

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2 years ago

Please answer I will mark brainliest

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3,466.32

Step-by-step explanation:

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26. The two triangles below are similar. What<br> is the perimeter of triangle XYZ?

AveGali [126]

Answer:

Step-by-step explanation:

6 ÷ 15 = 0.4

X to Y = 20 × 0.4 = 8

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2 years ago

If f and g are continuous functions with f(0) = 2 and the following limit, find g(0). lim_(x->0)[2 f(x) - g(x)] = -3

ladessa [460]

We have by distributivity of limits over sums, and continuity of $f(x)$ and $g(x)$ , that

$\displaystyle \lim_{x\to0} \bigg(2f(x) - g(x)\bigg) = 2 \lim_{x\to0} f(x) - \lim_{x\to0} g(x) = 2 f(0) - g(0)$

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1 year ago

The exponential distribution applies to lifetimes of a certain component. Its failure rate is unknown. Find the probability that

wlad13 [49]

Answer:

a) 0.0821

b) 0.0111

c) 0.0041

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \lambda e^{-\lambda x}$

In which $\lambda$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\lambda x}$

Either it lasts more 5 years or less, or it survives more than 5 years. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

In all 3 cases, we want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = 1 - e^{-5\lambda}$

(a) lambda=.5

$P(X \leq 5) = 1 - e^{-5*0.5} = 0.9179$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9179 = 0.0821$

(b) lambda=0.9

$P(X \leq 5) = 1 - e^{-5*0.9} = 0.9889$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9889 = 0.0111$

(c) lambda=1.1

$P(X \leq 5) = 1 - e^{-5*1.1} = 0.9959$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9959 = 0.0041$

7 0

3 years ago