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dimaraw [331]
2 years ago
12

The domain of F(x) is the set of all numbers greater then or equal to 0 and less or equal to 2.

Mathematics
1 answer:
lisov135 [29]2 years ago
5 0
The answer should be B false
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Keisha drank the average amount of water 1 of the days.

if he drinks 1 quart, his average is 21/13.

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What kind of number does distance always represent?
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Step-by-step explanation:There are other kinds of numbers that can be graphed on the number line, too. ... which are natural numbers and which are whole numbers is to think of a “hole,” which can be represented by 0. ... Notice that distance is always positive or 0.

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3 years ago
When 3+4i2+i is expressed in the form a+bi, what is the value of a?
mart [117]
Multiplying and dividing with conjugate of 2+i in the given expression, we get(3+ 4i)(2-i)/(2+i)(2-i)=(6-3i+8i-4i^2)/(4-i^2)=(10+5i)/(4+1)=(10+5i)/5=2+i=a+bi<span>Thus a=2</span>
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2 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
(4 + 18 ➗3) x 4 - 3 evaluate each expression
erastova [34]

Answer:

Step-by-step explanation:

(4 + 6) x 4 - 3

10 x 4 -3

40 -3

37

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