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natali 33 [55]
3 years ago
6

A keycode must contain 2 letters and 3 numbers. The letters may be any letter of the alphabet. The numbers should be any number

from 0 to 9. How many different keycode combinations are there?
Mathematics
2 answers:
OverLord2011 [107]3 years ago
8 0
The English alphabet contains 26 letters (a, b, c, ...y, z). 

The digits from 0 to 9 are a total of 10.


A keycode contains 2 letters, and 3 numbers, for example:

AB 598;  MM 139;    NT 498; ...


So there are 26 possible choices for the first letter, which can combined with any of the 26 possible choices for the second letter, so there are a total of 

26*26=676 possible pairs of letters.


Similarly, the 10 possible choices for the first number, which can be combined with the 10 possible choices for the second number, and the 10 possible choices for the third number make a total of :

10*10*10=1,000 possible triples of numbers.


Any of the 676 possible pairs of letters can be combined with any of the possible 1,000 triples of numbers. This makes a total of 

676*1,000=676,000 keycodes.


Answer: 676,000
professor190 [17]3 years ago
7 0

Answer:

There is 676,000 combinations

Step-by-step explanation:

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Suppose a large shipment of microwave ovens contained 12% defectives. If a sample of size 474 is selected, what is the probabili
melisa1 [442]

Answer:

P(\hat p>0.14)

And using the z score given by:

z = \frac{\hat p -\mu_p}{\sigma_p}

Where:

\mu_{\hat p} = 0.12

\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149

If we find the z score for \hat p =0.14 we got:

z = \frac{0.14-0.12}{0.0149}= 1.340

So we want to find this probability:

P(z>1.340)

And using the complement rule and the normal standard distribution and excel we got:

P(Z>1.340) = 1-P(Z

Step-by-step explanation:

For this case we have the proportion of interest given p =0.12. And we have a sample size selected n = 474

The distribution of \hat p is given by:

\hat p \sim N (p , \sqrt{\frac{p(1-p)}{n}})

We want to find this probability:

P(\hat p>0.14)

And using the z score given by:

z = \frac{\hat p -\mu_p}{\sigma_p}

Where:

\mu_{\hat p} = 0.12

\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149

If we find the z score for \hat p =0.14 we got:

z = \frac{0.14-0.12}{0.0149}= 1.340

So we want to find this probability:

P(z>1.340)

And using the complement rule and the normal standard distribution and excel we got:

P(Z>1.340) = 1-P(Z

4 0
3 years ago
Please help me
nikklg [1K]

Answer:

2. $1,100 in clothing sales gives Rachel a $260 paycheck

3 $.2,200 in clothing sales gives Rachel a $370 paycheck

5. $10,800 in clothing sales gives Rachel a 1,230 paycheck

Step-by-step explanation:

Here is the complete question :

Rachel works for a clothing store. She is $150 paid every week plus a 10% commission on any clothes that she sells. Which of the following is true in regards to Rachel's weekly paycheck? Chose three correct answers.

1. $100 in clothing sales gives Rachel a $250 paycheck

2. $1,100 in clothing sales gives Rachel a $260 paycheck

3 $.2,200 in clothing sales gives Rachel a $370 paycheck

4. $ 4,200 in clothing sales gives Rachel a $420 paycheck

5. $10,800 in clothing sales gives Rachel a 1,230 paycheck

6. $ 16,000 in clothing sales gives Rachel a 1,450 paycheck

The equation representing Rachel's paycheck = amount she earns every month + commission

If she makes  $100 in clothing sales, her paycheck = $150 + (0.1 x 100) = $160

If she makes  $1100 in clothing sales, her paycheck = $150 + (0.1 x 1100) = $260

If she makes  $2200 in clothing sales, her paycheck = $150 + (0.1 x 2200) = $370

If she makes  $4200 in clothing sales, her paycheck = $150 + (0.1 x 4200) = $570

If she makes  $10800 in clothing sales, her paycheck = $150 + (0.1 x 10800) = $1230

If she makes  $16000 in clothing sales, her paycheck = $150 + (0.1 x 16000) = $1750

thus the correct options are 2,3,5

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How many ten thousands in 2,689,456
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There are eight, if I'm understanding the question correctly. 8 is in the ten thousands place, at least
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A kite is fllying 115 ft above the ground. The length of the string to the kite is 150 ft, measured from the ground. Find the an
Kitty [74]

Answer:

The answer is sin θ = 115/150

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Answer:

Education in Nepal was long based on home-schooling and gurukulas.[citation needed] The first formal school (Durbar School), established by Jung Bahadur Rana in 1853, was intended for the elite. The birth of Nepalese democracy in 1951 opened its classrooms to a more diverse population.[citation needed] Education in Nepal from the primary school to the university level has been modeled from the very inception on the Indian system, which is in turn the legacy of the old British Raj.[1]

Step-by-step explanation:

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