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Alexus [3.1K]
3 years ago
15

Julie's MP3 player contains 860 songs. If 20% of the songs are rap songs and 15% of the songs are R&B songs, how many of the

songs are other types of songs?
Mathematics
2 answers:
docker41 [41]3 years ago
5 0
Total number of songs in Julie's MP3 player = 860
Percentage of rap songs in Julie's MP3 player = 20%
Then
Number of rap songs in Julie's MP3 player = (20/100) * 860
                                                                     = 860/5
                                                                     = 172
Percentage of R&B songs in Julie's MP3 player = 15%
Then
Number of R&B songs in Julie's MP3 player = (15/100) * 860
                                                                        = 1290/10
                                                                         = 129
So
The number of other types of songs in Julie's MP3 player = 860 - (172 + 129)
                                                                                             = 860 - 301
                                                                                             = 559
So the number of other types of songs in Julie's MP3 player was 559.
NNADVOKAT [17]3 years ago
4 0

Answer:

The number of other types of songs in Julie's MP3 player was 559.

Step-by-step explanation:

Total number of songs in Julie's MP3 player = 860

<u>Rap</u>

The percentage of rap songs in Julie's MP3 player = 20%

The Number of rap songs is 172

<u>R&B</u>

Percentage of R&B songs in Julie's MP3 player = 15%

The Number of R&B songs in Julie's MP3 player = 129

<u>Other Music</u>

The number of other types of songs in Julie's MP3 player = 559

So the number of other types of songs in Julie's MP3 player was 559.

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The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event th
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Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

f_x(x) = \dfrac{1}{1500} ; o \le x \le   \  1500

The function of the insurance is:

I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \  \ \ \ 250 \le x \le 1500}} \right.

Hence, the variance of the insurance can also be an account forum.

Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2

here;

E[I(x)] = \int f_x(x) I (x) \ sx

E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx

= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}

\dfrac{5}{12} \times 1250

Similarly;

E[I^2(x)] = \int f_x(x) I^2 (x) \ sx

E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx

= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}

\dfrac{5}{18} \times 1250^2

∴

Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]

Finally, the standard deviation  of the insurance payment is:

= \sqrt{Var(I(x))}

= 1250 \sqrt{\dfrac{5}{48}}

≅ 404

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igomit [66]

Answer:

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Answer:

19.6 cubic centimeters

Step-by-step explanation:

The ratio of two similar 3D figures' volumes is equivalent to the ratio of their corresponding lengths (whether it's width, height, length, radius, etc) cubed.

Here, we know the cones are similar, and we know the smaller radius is 2 and the larger radius is 3. Then the ratio of the volume of the smaller one to the volume of the larger one is (2/3)^3=8/27.

We also know the actual volume of the larger cone is 66. So, if x is the volume of the smaller cone, we have the proportion:

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Cross-multiply:

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Divide by 27:

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<em>~ an aesthetics lover</em>

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