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2 years ago

7

20 POINTS PLEASE HELP

1 answer:

Andru [333]2 years ago

8 0

The figure RQTS is a clockwise rotation of figure DEFC, then:
The image of CD is
Answer: Option C. RS

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Please explain how do i fill in the blank and how do i get the answer

polet [3.4K]

9/343

is the correct answer

mark me as the brainliest

5 0

2 years ago

Read 2 more answers

Integrate sin²2x cos²2xdx

Masteriza [31]

$\sin^2x=\dfrac{1-\cos2x}2$
$\cos^2x=\dfrac{1+\cos2x}2$

From the identities above, you have

$\sin^22x\cos^22x=\dfrac{(1-\cos4x)(1+\cos4x)}4=\dfrac{1-\cos^24x}4$

Applying once more, you have

$\dfrac{1-\cos^24x}4=\dfrac{1-\dfrac{1+\cos8x}2}4=\dfrac{1-\cos8x}8$

So,

$\displaystyle\int\sin^22x\cos^22x\,\mathrm dx=\frac18\int(1-\cos8x)\,\mathrm dx=\frac x8-\frac1{64}\sin8x+C$

3 0

2 years ago

Evaluate 3 3/4 ÷ 5/6 =

jeka94

Answer:

9/2 or 4.5

Step-by-step explanation:

15/4x6/5 = 3/2x3 = 9/2

6 0

2 years ago

Please find x.............

emmasim [6.3K]

Answer:

$\underline{\boxed{ x=50}}$

Step-by-step explanation:

$164 = x + 114 \\ x = 164 - 114 \\ x =50$

4 0

2 years 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