Answer:13
Step-by-step explanation: Your going up on the number line.
Hope I was able to help! :)
Answer:

Step-by-step explanation:
Unshaded Right Triangle

Whole Triangle

Shaded triangle

If you dont want as many steps, just calculate the triangle and dont worry about the right triangle the formula is

Answer:
From the central limit theorem we know that the distribution for the sample mean
is given by:

And now for the deviation we have this:

So then the correct answer for this caee would be:
c. 1.30 ounces.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
From the central limit theorem we know that the distribution for the sample mean
is given by:

And now for the deviation we have this:

So then the correct answer for this caee would be:
c. 1.30 ounces.
Answer:
At least 401 chosen guests
Step-by-step explanation:
Represent those with same birth day with Same
So:

Represent number of people with n
There are 365 days in a year and 364 days out of these days is not your birthday.
So, there the probability that n people do not share your birthday is 
i.e.

Solving further, we have that:



We're calculating the probability that
.
So, we have:

Collect Like Terms




Take natural logarithm (ln) of both sides


Apply laws of logarithm

Multiply through by -1

Solve for n

Reorder




(approximated)
This implies that n = 401, 402, 403 .....
i.e

So, at least 401 people has to be invited
Hey sorry I can’t help but use Mathaway or Photomath to help!