Answer:
23
Step-by-step explanation:
you subtract 50 the max number from 27 the Lowest number
By solving a system of equations, we will see that you must use 6.67 pounds of chocolate-covered almonds.
<h3>
How to solve the system of equations?</h3>
First, we need to define the two variables:
- x = number of pounds of the $9.99 chocolate-covered almonds used.
- y = number of pounds of the $5.99 chocolate-covered raisins used.
If we mix that, we will get:
x*$9.99 + y*$5.99 = (x + y)*C
That equation means that the cost of the individual elements of the mixture, must be equal to the total cost of the mixture. Where C is the cost of each pound of the mixture, and we know that:
C = $8.75
y = 3
Replacing that, we get:
x*$9.99 + 3*$5.99 = (x + 3)*$8.75
Now we can solve that for x:
x*$9.99 - x*$8.75 = 3*$8.75 - 3*$5.99
x*$1.24 = $8.28
x = $8.28/$1.24 = 6.67
Then you must use 6.67 pounds of the chocolate-covered almonds.
If you want to learn more about systems of equations:
brainly.com/question/13729904
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Answer:
0.1426 = 14.26% probability that at least one of the births results in a defect.
Step-by-step explanation:
For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability that a birth results in a defect is independent of any other birth. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).
This means that 
A local hospital randomly selects five births.
This means that 
What is the probability that at least one of the births results in a defect?
This is:

In which



0.1426 = 14.26% probability that at least one of the births results in a defect.