We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.
The probability of a data score between two z-scores is given by formula 
.
Using above formula, we will get:
Now we will use normal distribution table to find probability corresponding to both z-scores as:
Now we will convert 
into percentage as:
Upon rounding to nearest tenth of percent, we will get:
Therefore, our required probability is 27.4% and option C is the correct choice.
Answer:
Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.
Step-by-step explanation:
We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.
<em>Let X = incomes for the industry</em>
So, X ~ N(
)
Now, the z score probability distribution is given by;
Z = 
~ N(0,1)
where, 
= mean income of firms in the industry = 95 million dollars
= standard deviation = 5 million dollars
So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)
P(X < 100) = P( < 
) = P(Z < 1) = 0.8413 {using z table]
Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.
Answer:
7
Step-by-step explanation:
(15x + 5) and (-2 + 16x) represents the measures of corresponding angles.
Measures of corresponding angles are equal.
Therefore,
- 2 + 16x = 15x + 5
16x - 15x = 5 + 2
x = 7
Huh I 8 so yeah bye nice talking to ya
Answer:
-30bc + 50bd
Step-by-step explanation:
3a + 5b/3c - 5d = 3a - 5b/3c - 5d
(3c - 5d)(3a - 5b) = (3a + 5b)(3c - 5d)
9ca - 15bc - 15da + 25db = 9ac - 15ad + 15bc - 25bd
9ca -15bc - 15da + 25db - 9ac + 15ad - 15bc + 25bd
9ca and -9ac and -15da and 15ad cancel each other out, so
-30bc + 50db is left
