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zmey [24]
3 years ago
11

The area of a circle depends on the radius. Use the equation A(r) = (3.1415..)r^2 to compute A(6). Round your answer to the near

est hundredth.
Mathematics
1 answer:
fredd [130]3 years ago
3 0
Hello!

To solve this you put 6 in for r

When you put this into a calculator you get the answer of 113.10

Hope this helps!
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Angel started out with $20 in his bank . He had been adding $30 per month . nicole started with no money in her bank account but
lana66690 [7]

Answer:

2 months

Step-by-step explanation:

started with 20, an added 30 each month so 30 plus 30 equals 60, 2 months add it to 20, 60 plus 20 is 80. other girl started with nothing and added 40 each month so 40 plus 40 equals 80. also 2 months

8 0
3 years ago
Solve the equation.<br><br> 4(x – 6) = 16<br> A. 8<br> B. 10<br> C. –10<br> D. 12
vladimir2022 [97]

Answer:

a.

Step-by-step explanation:

its 8 hope this helps!

8 0
3 years ago
Read 2 more answers
An aquarium is on sale for $59.50. If this price represents a 15% discount from the original price, what is the original price t
Lina20 [59]
x-\ original\ price\\\\&#10;x+15\%x=59,5\\\\&#10;x-0,15x=59,5\\\\&#10;0,85x=59,5\ \ \ | divide\ by\ 0,85\\\\&#10;x=70\$\\\\Original\ price\ is\ 70\$.
4 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
Ms Amani and mr Blake each ordered supplies for their classroom
swat32

Could you please write the full question so we can correctly answer the question for you? Or upload a picture of the question, if you don’t want to type?

Hope this helps!

6 0
3 years ago
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