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Luda [366]
2 years ago
11

What percent of a dollar is a nickel and a penny? It is ______% of a dollar.

Mathematics
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

6 %

Step-by-step explana

100 pennies makes a dollar... you just add 5 for the nickel and 1 fo the penny and it makes 6

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1/2

Step-by-step explanation:

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3 years ago
Help please easy math question
Rus_ich [418]

Answer:

4.8m

Step-by-step explanation:

3 0
3 years ago
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Your task, to replicate this musical frequency is to write a method harmonicSeries that accepts an integer parameter n and retur
LUCKY_DIMON [66]

Answer:

Step-by-step explanation:

JUst try your best

5 0
3 years ago
According to a recent​ survey, the average daily rate for a luxury hotel is ​$239.67. Assume the daily rate follows a normal pro
xenn [34]

Answer:

a) 0.8000

b) 0.1080

c) 0.3260

d) $269.1

Step-by-step explanation:

Mean = xbar = $239.67

standard deviation = $22.93

a) The probability that a randomly selected luxury​ hotel's daily rate will be less than $259 = P(x < 259)

We need to standardize the $259 in z-score.

The standardized z-score is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (259 - 239.67)/22.93 = 0.843

To determine the probability that a randomly selected luxury​ hotel's daily rate will be less than $259

P(x < 259) = P(z < 0.843)

We'll use data from the normal probability table for these probabilities

P(x < 259) = P(z < 0.843) = 1 - P(z ≥ 0.843) = 1 - P(z ≤ - 0.843) = 1 - 0.2 = 0.8000

b) The probability that a randomly selected luxury​ hotel's daily rate will be more than $268 = P(x > 268)

We need to standardize the $268 in z-score.

z = (x - xbar)/σ = (268 - 239.67)/22.93 = 1.235

To determine the probability that a randomly selected luxury​ hotel's daily rate will be more than $268

P(x > 268) = P(z > 1.235)

We'll use data from the normal probability table for these probabilities

P(x > 268) = P(z > 1.235) = 1 - P(z ≤ 1.235) = 1 - 0.892 = 0.1080

c) The probability that a randomly selected luxury​ hotel's daily rate will be between $236 and $256 = P(236 < x < 256)

We need to standardize the $236 and $256 in z-score.

z = (x - xbar)/σ = (256 - 239.67)/22.93 = 0.712

z = (x - xbar)/σ = (236 - 239.67)/22.93 = - 0.16

To determine the probability that a randomly selected luxury​ hotel's daily rate will be between $236 and $256

P(236 < x < 256) = P(-0.16 < z < 0.712)

We'll use data from the normal probability table for these probabilities

P(236 < x < 256) = P(-0.16 < z < 0.712) = P(z < 0.712) - P(z < -0.16) = [1 - P(z ≥ 0.712)] - [1 - P(z ≥ -0.16)] = [1 - P(z ≤ -0.712)] - [1 - P(z ≤ 0.16)] = (1 - 0.238) - (1 - 0.564) = 0.762 - 0.436 = 0.3260

d) The managers of a local luxury hotel would like to set the​ hotel's average daily rate at the 90th​percentile, which is the rate below which 90​% of​ hotels' rates are set. What rate should they choose for their​ hotel?

We need to obtain the z' value that corresponds to P(z ≤ z') = 0.90

From the normal distribution table,

z' = 1.282

z' = (x - xbar)/σ

1.282 = (x - 239.67)/22.93

x = (1.282 × 22.93) + 239.67 = $269.1

7 0
3 years ago
Solve for x -x2+3x+4=0
Maksim231197 [3]

The correct answer is D) x = 4 and x = -1

In order to solve this, we plug into the quadratic equation.

x = \frac{-b +/- \sqrt{b^{2} - 4ac } }{2a}

In this case, a is the coefficient of x^2 (-1), b is the coefficient of x (3) and c is the constant (4). Now we plug in and solve.

x =  \frac{-(3) +/- \sqrt{3^{2} - 4(-1)(4) } }{2(-1)}

x =  \frac{-3 +/- \sqrt{9 + 16 } }{-2}

x = \frac{-3 +/- \sqrt{25 } }{-2}

x =  \frac{-3 +/- 5 } }{-2}

And then we'll split into the two possible answers.

(-3 + -5)/-2 = -8/-2 = 4

(-3 - -5)/-2 = 2/-2 = -1

3 0
3 years ago
Read 2 more answers
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