Answer:
0.57142
Step-by-step explanation:
A normal random variable with mean and standard deviation both equal to 10 degrees Celsius. What is the probability that the temperature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit?
We are told that the Mean and Standard deviation = 10°C
We convert to Fahrenheit
(10°C × 9/5) + 32 = 50°F
Hence, we solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 59 °F
μ is the population mean = 50 °F
σ is the population standard deviation = 50 °F
z = 59 - 50/50
z = 0.18
Probability value from Z-Table:
P(x ≤59) = 0.57142
The probability that the temperature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit
is 0.57142
Multiply the dimensions by 1.5 and Pamela's suitcase becomes 42" x 24" x 12".
<span>Volume of 28" x 16" x 8": 3584 in^3 </span>
<span>Volume of 42" x 24" x 12": 12,096 in^3 </span>
<span>The volume ratio is 3.375, and yes, you have to cube it.</span>
Answer:
$104.70
Step-by-step explanation:
The equation would be set up like this: 80.45 + 20.50(3) - 37.25. You started off with $80.45 in your bank account and deposited, or added, $20.50 every day on Tuesday, Wednesday and Thursday. That would mean you added $20.50 three times. Adding $80.45 + $20.50 + $20.50 + $20.50, simplified to $80.45 + $20.50(3) would get you $141.95 in total. Then, on Friday, you withdraw $37.25, getting the equation $141.95 - $37.25, leaving $104.70 for the weekend.
Given:
Temperature in the morning = 6°F.
By the late afternoon, the temperature had dropped 9°F.
To find:
The temperature by the late afternoon.
Solution:
Temperature by the late afternoon = Morning temperature - Dropped temperature
Using the given values and the above formula, we get
Temperature by the late afternoon = 6°F - 9°F
Temperature by the late afternoon = -3°F
Therefore, the temperature by the late afternoon is -3°F.
Answer:
how do we graph on here
Step-by-step explanation:
