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3 years ago

A two-digit locker combination is made up of nonzero digits and no digit is repeated in any combination.

1 answer:

3 0

$|\Omega|=9\cdot8=72\\
A\cap B=\{(1,2),(1,4),(1,6),(1,8)\}\\
|A\cap B|=4\\\\
P(A\cap B)=\dfrac{4}{72}=\dfrac{1}{18}\Rightarrow \text{A}$

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A video game enthusiast saved $750 to spend on a video game player and games. The player costs $400. The games costs $49 each. A

user100 [1]

Answer:

7 games

Step-by-step explanation:

750 (have) - 400 (game system) = 350 (left to spend)

divide 350 by 49 and get 7 games

you could also round 49 up to 50 to simplify the division.

4 0

4 years ago

Whatis the tigonemtric form of -3+4i

Anettt [7]

We calculate the module:

$|z|\: = \: \sqrt{( - 3) ^{2} \: + \: {4}^{2} }$

$|z| \: = \: \sqrt{9 \: + \: 16}$

$|z| \: = \: \sqrt{25}$

$\boxed{|z| \: = \: 5}$

We calculate the angle formed by "z":

$\arctan( \frac{4}{ - 3} ) \: = \: \underline{0.92729521 \: \text{rad}}$

We pass it to degrees:

$0.92729521 \: \times \: \frac{180}{\pi}$

$\frac{166.91313924}{\pi}$

$\frac{166.91313924}{3.14159265}$

$\boxed{ 53.13°}$

Now we use this formula to transform it into a trigonometric form:

$\boxed{z \: = \: |z| \times \: ( \cos( \alpha) \: + \: i \: \times \: \sin( \alpha))}$

<h3>We substitute the values already obtained:</h3>

$\boxed{ \bold{z \: = \: 5 \times \: ( \cos( 53.13°) \: + \: i \: \times \: \sin( 53.13°))}}$

<h2>Answer: </h2>

$\boxed{ \bold{z \: = \: 5 \times \: ( \cos( 53.13°) \: + \: i \: \times \: \sin( 53.13°))}}$

<h3><em><u>MissSpanish</u></em> </h3>

7 0

3 years ago

Suppose we want to choose 5 objects, without replacement, from 15 distinct objects.

vovangra [49]

The number of ways it can be done if order of the choices is taken into consideration is 15,444 ways

<h3>Permutation and combination</h3>

Permutation has to do with arrangement and combination has to do with selection.

If we are choosing 5 objects, without replacement, from 15 distinct objects, this has to do with permutation if the order of the choices is taken into consideration

15P5 = 15!/(15-5)!5!
15P5 = 15!/10!5!
15P5 = 15*14*13*12*11/5*4*3*2

15P5 = 15,444 ways

Hence the number of ways it can be done if order of the choices is taken into consideration is 15,444 ways

Learn more on permutation here: brainly.com/question/11732255

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Complete question

<em>Suppose we want to choose 5 objects, without replacement, from 15 distinct objects. How many ways can this be done, if the order of the choices is taken into consideration?</em>

6 0

2 years ago

A magnetic tape storing information in binary form has been corrupted, so it can only be readwith bit errors. The probability th

Natali [406]

Answer:

6/7 ≈ 85.7%

Step-by-step explanation:

The probability that a 0 was transmitted and received is 0.5·0.9 = 0.45.

The probability that a 1 was transmitted and a 0 received is 0.5·0.15 = 0.075.

Then the probability that a 0 was received is 0.45 +0.075 = 0.525.

The probability of interest is ...

p(0 transmitted | 0 received) = p(0 transmitted & received)/p(0 received)

= 0.45/0.525 = 6/7 ≈ 85.7%

8 0

4 years ago

Record the product use expanded form to help 6x 532

gogolik [260]

6×500=3000
6×30=180
6×2=12
3000+180+12=3182
3182 is the final answer

7 0

3 years ago

Read 2 more answers