The number of marshmallows an adult can fit in their mouth is skewed right with a mean of 6.5 and a standard deviation of 0.58. What is the probability that a random sample of 40 adults would have a mean of at least 7
marshmallows?
approximately 0
0.1943
0.2934
0.5367

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

∠cbe and ∠deb are same-side interior angles.

i took the test

$3,200 x 7.25% = $3,200 x .0725 = $232 interest monthly

$232 monthly x 12 annual = $2, 784

step-by-step explanation:

step-by-step explanation: