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pishuonlain [190]
2 years ago
8

You want to make a necklace with 5 beads on it; you have a bag that contains 5 differently colored beads. You draw a bead from t

he bag and string it on the necklace. Then you draw another bead and add it to the necklace. This process is repeated until all of the beads are used. How many different color combinations are possible for the necklace?
Mathematics
1 answer:
Vladimir [108]2 years ago
8 0
There are 120 different combinations.

There are 5 to use for the first bead; then after it is used, 4 for the second; 3 for the third; 2 for the fourth; and 1 for the fifth and last bead:

5*4*3*2*1 = 120.
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PLEASE HELP/ANSWER! From 1980 to 1990, Lior’s weight increased by 25%. If his weight was k kilograms in 1990, what was it in 198
Elena-2011 [213]

The weight in 1980 is \frac{4k}{5} kilograms

<em><u>Solution:</u></em>

From 1980 to 1990, Lior’s weight increased by 25%

His weight is "k" kilograms in 1990

<em><u>To find: weight in 1980</u></em>

This is a percentage increase problem

Let "x" be the weight in kilograms in 1980

<em><u>The percentage increase is given by formula:</u></em>

\text{Percentage increase } = \frac{\text{Final value - initial value}}{\text{initial value}} \times 100

Here,

Initial value in 1980 = x

Final value in 1990 = k

Percentage increase = 25 %

<em><u>Substituting the values in formula,</u></em>

25 = \frac{k-x}{x} \times 100\\\\25x = 100(k-x)\\\\x = 4(k-x)\\\\x = 4k - 4x\\\\5x = 4k\\\\x = \frac{4k}{5}

Thus the weight in kilograms in 1980 is \frac{4k}{5}

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Every 3 weeks 8 students will lose their metro cards... How many will lose it in 9 weeks
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ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

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(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

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2 years ago
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Answer:

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