Answer:
Step-by-step explanation:
The number of samples is large(greater than or equal to 30). According to the central limit theorem, as the sample size increases, the distribution tends towards normal. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 22199
σ = 5300
n = 30
the probability that a senior owes a mean of more than $20,200 is expressed as
P(x > 20200)
Where x is a random variable representing the average credit card debt for college seniors.
For n = 30,
z = (20200 - 22199)/(5300/√30) =
- 2.07
Looking at the normal distribution table, the probability corresponding to the z score is 0.0197
P(x > 20200) = 0.0197
The answer is b
multiply 5.32 times 46 and then 6.01 times 46 and add
or it could be a if you are rounding
<span>The characteristic equation for this is r^2 + yr + 1 = 0. Its roots are r = y/2 +/- isqrt(1-y^2/4). Finding the general solution to this equation gives us y = sqrt(20/9), which is the value of the damping coefficient y for which the quasi period of the damped motion is 50% greater.</span>
Question:
Jim has 19,700 g of sand in his sandbox. He brings home another 6,300 g of sand from the beach to add to his. How many kilograms of sand does Jim have in his sandbox now?
Answer:
The amount of kilogram of sand Jim now has in his sand box is 26.00 kg
Step-by-step explanation:
The question relates to the method of addition or adding numbers.
Here, we note that the initial mass of sand in the sandbox = 19700 g
Added mass of sand to the sandbox = 6,300 g
Therefor the total mass of sand in the sand box is given by;
19,700 + 6,300 = 26,000 g
Therefore, the amount of kilogram of sand Jim now has in his sand box is 26,000 g which is equal to 26000/1000 = 26.00 kg.