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Mashutka [201]
3 years ago
13

At a school fair, 70% of the people are under 16 years old. One-third of the people remaining are teachers. If there are 21 teac

hers at the fair, how many people are there at the fair total? A. 63, B. 126, C. 189, D. 210. Please I need help.
Mathematics
1 answer:
Free_Kalibri [48]3 years ago
7 0
So, "one third of remaing" is 21, this means that the "remaining" was three times as much: 21*3=63

This 63 is the 30 % of all the people, since they are the remaining of 100-70%.

so: 63=30/100x

we now multiply both sides by 100:

6300=30x

and divide by 30:
210

so all the people at the fair numbered 210!

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Does anyone know the answer to this ?
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Answer:answer b

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Assume that adult IQ scores on the Weschler IQ test are normally distributed with a mean of 100 points and a standard deviation
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Hi there!

With a normal distribution, we can use the operation 'normalcdf' on a calculator to find the probability that a randomly selected adult has an IQ between 96 and 111.

Here is the format for using the operation:

<h2>normalcdf(LB, UB, μ, σ) </h2>

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License plate numbers in a certain state consists of seven characters. The first character is a non-zero digit (1 through 9). Th
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Answer:

a)  333,135,504 different plates

b) 230,315,904 different plates

c) 180,835,200 different plates

Step-by-step explanation:

Pattern: Digit(1-9)-Letter-Letter-Letter-Letter-Digit(1-9)-Digit (1-9)

We will calculate the number of possibilities for the digits part, then for the letters part, then we'll multiply them together.

For the digits, we have 3 numbers, first and last 2 positions. We can consider this is a single 3-digit number, where n = 9 (since they are non-zero digits) and r = 3.  

For the letters part, it's basically a 4-letter word, where n = 26 (A through Z) and r = 4.

(a) How many different license plate numbers are possible?

No limitation on repeats for this question:

For the digits, we have 9 * 9 * 9 = 729 (since repetition is allowed, and we can pick any digit from 0 to 9 for each position)

For the letters we have: 26 * 26 * 26 * 26 = 456,976

Because the digits and letters arrangements are independent from each other, we multiply the two numbers of possibilities to have the global number of possibilities:

P = 729 * 456976 = 333,135,504 different plates, when there's no repeat limitation.

(b) How man license plate numbers are possible if no digit appears more than once?

Repeats limitation on digits:

For the digits, we have 9 * 8 * 7 = 504 (since repetition is NOT allowed, we can pick any of 9 digits for first position, then any 8 remaining and finally any 7 remaining at the end)

For the letters we still have: 26 * 26 * 26 * 26 = 456,976

Because the digits and letters arrangements are independent from each other, we multiply the two numbers of possibilities to have the global number of possibilities:

P = 504 * 456976 = 230,315,904 different plates, when there's no repeat on the digits.

(c) How man license plate numbers are possible if no digit or letter appears more than once?

Repeats limitation on both digits and letters:

For the digits, we have 9 * 8 * 7 = 504 (

For the letters we still have: 26 * 25 * 24 * 23 = 358,800

Because the digits and letters arrangements are independent from each other, we multiply the two numbers of possibilities to have the global number of possibilities:

P = 504 * 358800 = 180,835,200 different plates, when there's no repeat on the digits AND on the letters.

6 0
4 years ago
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