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vodomira [7]
3 years ago
14

Said salió a comer alitas con sus amigos. Pidieron 2 promociones de 25 alitas por $189, pero acordaron que cada uno pagaría la c

antidad de alitas que comiera. Si Said comió 12 alitas, ¿cuánto pagó?
Mathematics
1 answer:
vichka [17]3 years ago
3 0

Answer:

$45.36

Step-by-step explanation:

De acuerdo al enunciado, 2 promociones de 25 alitas cuestan $189. Esto significa que por ese valor compraron 50 alitas.

$189/50= $3.78

Cada alita cuesta $3.78 y para encontrar el valor que pagó Salid, tienes que multiplicar el precio unitario por 12:

$3.78*12= $45.36

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A gardener is planting two types of trees: Type A is 4 feet tall and grows at a rate of 24 inches per year. Type B is 5 feet tal
goldenfox [79]
It would take 5 yrs. for both trees to be the same height.
4 0
3 years ago
The Empirical Rule The following data represent the length of eruption for a random sample of eruptions at the Old Faithful geys
ad-work [718]

Answer:

(a) Sample Standard Deviation approximately to the nearest whole number = 6

(b) The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is.

(c) The percentage of eruptions that last between 92 and 116 seconds using the empirical rule is 95%

(d) The actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) The percentage of eruptions that last less than 98 seconds using the empirical rule is 16%

(f) The actual percentage of eruptions that last less than 98 seconds is 15.866%

Step-by-step explanation:

(a) Determine the sample standard deviation length of eruption.

Express your answer rounded to the nearest whole number.

Step 1

We find the Mean.

Mean = Sum of Terms/Number of Terms

= 90+ 90+ 92+94+ 95+99+99+100+100, 101+ 101+ 101+101+ 102+102+ 102+103+103+ 103+103+103+ 104+ 104+104+105+105+105+ 106+106+107+108+108+108 + 109+ 109+ 110+ 110+110+110+ 110+ 111+ 113+ 116+120/44

= 4582/44

= 104.1363636

Step 2

Sample Standard deviation = √(x - Mean)²/n - 1

=√( 90 - 104.1363636)²+ (90-104.1363636)² + (92 -104.1363636)² ..........)/44 - 1

= √(199.836777 + 199.836777 + 147.2913224+ 102.7458678+ 83.47314049+ 26.3822314+ 26.3822314+ 17.10950413+17.10950413+ 9.836776857+ 9.836776857, 9.836776857+9.836776857+ 4.564049585+ 4.564049585+ 4.564049585+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 0.01859504133+ 0.01859504133+ 0.01859504133+ 0.7458677685+ 0.7458677685+ 0.7458677685+ 3.473140497+ 3.473140497+ 8.200413225+ 14.92768595+ 14.92768595+ 14.92768595+ 23.65495868+ 23.65495868+ 34.38223141+ 34.38223141+34.38223141+ 34.38223141+ 34.38223141+47.10950414+ 78.56404959+ 140.7458677+ 251.6549586) /43

= √1679.181818/43

= √39.05073996

= 6.249059126

Approximately to the nearest whole number:

Mean = 104

Standard deviation = 6

(b) On the basis of the histogram drawn in Section 3.1, Problem 28, comment on the appropriateness of using the Empirical Rule to make any general statements about the length of eruptions.

The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is .

(c) Use the Empirical Rule to determine the percentage of eruptions that last between 92 and 116 seconds.

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

Mean = 104, Standard deviation = 6

For 68% μ - σ = 104 - 6 = 98, μ + σ = 104 + 6 = 110

For 95% μ – 2σ = 104 -2(6) = 104 - 12 = 92

μ + 2σ = 104 +2(6) = 104 + 12 = 116

Therefore, the percentage of eruptions that last between 92 and 116 seconds is 95%

(d) Determine the actual percentage of eruptions that last between 92 and 116 seconds, inclusive.

We solve for this using z score formula

The formula for calculating a z-score is is z = (x-μ)/σ

where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 104, Standard deviation = 6

For x = 92

z = 92 - 104/6

= -2

Probability value from Z-Table:

P(x = 92) = P(z = -2) = 0.02275

For x = 116

z = 92 - 116/6

= 2

Probability value from Z-Table:

P(x = 116) = P(z = 2) = 0.97725

The actual percentage of eruptions that last between 92 and 116 seconds

= P(x = 116) - P(x = 92)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage = 0.9545 × 100

= 95.45%

Therefore, the actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) Use the Empirical Rule to determine the percentage of eruptions that last less than 98 seconds

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

For 68% μ - σ = 104 - 6 = 98,

Therefore, 68% of eruptions that last for 98 seconds.

For less than 98 seconds which is the Left hand side of the distribution, it is calculated as

= 100 - 68/2

= 32/2

= 16%

Therefore, the percentage of eruptions that last less than 98 seconds is 16%

(f) Determine the actual percentage of eruptions that last less than 98 seconds.

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 98

Z score = x - μ/σ

= 98 - 104/6

= -1

Probability value from Z-Table:

P(x ≤ 98) = P(x < 98) = 0.15866

Converting to percentage =

0.15866 × 100

= 15.866%

Therefore, the actual percentage of eruptions that last less than 98 seconds is 15.866%

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A crepe recipe calls for 2 eggs, 1 cup of flour, and 1 cup of milk. How much flour would you need if you use 5 eggs?
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Answer: 2.5 cups of flour is needed for 5 eggs.

Step-by-step explanation:

The crepe recipe calls for 2 eggs, 1 cup of flour, and 1 cup of milk. This means that the ratio of eggs to cup of flour to cup of milk required is 2:1:1

Therefore, if you use 5 eggs, the proportions would increase by 2.5. Multiplying by 2.5, it becomes

5 : 2.5 : 2.5

Therefore, you would need 2.5 cups of flour for 5 eggs.

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Answer:

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a31 =  -22 + (31-1)3

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