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shusha [124]
2 years ago
14

Mal downloads 4 songs each week for 5 weeks.Then he sorts the songs into equal groups.There are 10 songs in each group.How many

groups are there?
Mathematics
2 answers:
12345 [234]2 years ago
8 0
4(5)= 20

20/10= 2

There are 2 groups of songs.

Hope this helps!
Artemon [7]2 years ago
7 0
There are 2 groups cause 4x5=20 then 20 divided by 2 is 10 so the answer is 2
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If y=1/2x+4, then find the equation of the line parallel through the point (6,3).
tatyana61 [14]

Answer:

x=2y

Step-by-step explanation:

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5 0
2 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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53 miles per hour because 265 divided 5 is 53.
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