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3 years ago

14

What is the answer to this: (2^2)^2/2*2^4

2 answers:

zalisa [80]3 years ago

4 0

Answer:

64

Step-by-step explanation:

sertanlavr [38]3 years ago

3 0

Answer:

64

Step-by-step explanation:

2^2*2^4

4*16

=64

hope it helps

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a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou

Fed [463]

Answer:

$\displaystyle P(A)=\frac{6}{11}$

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

$\Omega=\{WW,WN,NW,NN\}$

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

$A=\{WN,NW\}$

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

$\displaystyle \frac{5}{11}$

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

$\displaystyle \frac{6}{10}$

Thus the option WN has the probability

$\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}$

Now for the second option NW. The initial probability to pick a non-white counter is

$\displaystyle \frac{6}{11}$

The probability to pick a white counter is

$\displaystyle \frac{5}{10}$

Thus the option NW has the probability

$\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}$

The total probability of event A is the sum of both

$\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$

$\boxed{\displaystyle P(A)=\frac{6}{11}}$

7 0

2 years ago

Find the indefinite integral. (use c for the constant of integration.) sin3 5θ cos 5θ dθ

Elina [12.6K]

I'm guessing you mean

$\displaystyle\int\sin^35\theta\cos5\theta\,\mathrm d\theta$

Let $t=5\sin\theta$ . Then $\mathrm dt=5\cos5\theta\,\mathrm d\theta$ . So we have

$\displaystyle\int\sin^35\theta\cos5\theta\,\mathrm d\theta=\frac15\int\sin^35\theta(5\cos5\theta)\,\mathrm d\theta$

$=\displaystyle\frac15\int t^3\,\mathrm dt$

$=\dfrac{t^4}{20}+C$

$=\dfrac{\sin^4\5\theta}{20}+C$

7 0

3 years ago

Goran ran three times last week. Here are the distances he ran (in kilometers).9.94,6.65,10 What is the total distance Goran ran

kolbaska11 [484]

Answer:

26.59 kilometers

Step-by-step explanation:

7 0

2 years ago

Laura finished her history assignment in 1/4 hours. Then she completed her English assignment in 3/5 hours. What was the total a

kobusy [5.1K]

Answer:

0.85 hours

$\frac{17}{20}$ hours

Step-by-step explanation:

8 0

2 years ago

10 times as much as blank is 1/10 of 900

ludmilkaskok [199]

9 times 10 is 90.

90 is 1/10 of 900.

The answer is 9.

4 0

3 years ago