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3 years ago

6

The largest score in a distribution is 88 and the smallest is 14. You plan to construct a grouped distribution with 15 intervals

. What will be the size of your class interval?
(Statistics)

1 answer:

maks197457 [2]3 years ago

7 0

88 - 14 = 72

72/15 = 4.8 ≈ 5

Class interval = 5

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1. The C major key, starting with the middle C, consists of seven notes (white keys on the piano) with the following frequencies

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1.) <span>the ratio of the note A to middle C = 440.0 / 261.6 = 1.6820 ≈ 1.6818

2.) </span><span>the ratio of the note D to middle C = 293.7 / 261.6 = 1.1227

3.) </span><span>D# = 293.6 x 1.0595 = 311.1

4.) </span><span>the ratio of the frequency of G to C is 1 : 262/392 = 1 : 0.6683 ≈ 3 : 2

5.) </span><span>the ratio of the frequency of E to C = 1 : 262/330 = 1 : 0.7939 ≈ 5 : 4

6.) The </span><span>component in a musical note is referred to as the pitch of a sound is the frequency.</span>

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3 years ago

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mia wants to earn 2500 in her job at a bycicle store she is paid 55 for every bycicle she tunes up the minimum number of bycicle

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Answer: 46

Explanation: If Mia is paid $55 for every bicycle, she would need to fix up/sell 46 bicycles. This is because 45 x 55 = 2475 and Mia wants over 2500 but 46 x 55 = 2530. Therefore, 46 is the minimum number of bicycles.

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The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

<em>The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.</em>

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A circle has a circumference of 9 pi meters what is the area of this circle

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Answer: 63.62 square m.

Step-by-step explanation:

Find the Radius so 9/2= 4.5

Area = π r²

Area 3.14(20.25)

Area equles 63.62

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The teacher could get a whole pie and five individual slices but if not then...... The teacher can get one whole pie and cut each slice into 2 but there will be a remainder of 1 slice which the teacher can have......hope this helps

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