Answer:
Linear function
Step-by-step explanation:
4
<span><span> 4c2+6c-3c2-2c-3</span> </span>
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "c2" was replaced by "c^2". 1 more similar replacement(s).
Step by step solution :<span>Step 1 :
</span><span>Equation at the end of step 1 :</span><span><span> ((((4•(c2))+6c)-3c2)-2c)-3
</span><span> Step 2 :</span></span><span>Equation at the end of step 2 :</span><span> (((22c2 + 6c) - 3c2) - 2c) - 3
</span><span>Step 3 :</span>Trying to factor by splitting the middle term
<span> 3.1 </span> Factoring <span> c2+4c-3</span>
The first term is, <span> <span>c2</span> </span> its coefficient is <span> 1 </span>.
The middle term is, <span> +4c </span> its coefficient is <span> 4 </span>.
The last term, "the constant", is <span> -3 </span>
Step-1 : Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -3 = -3</span>
Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is <span> 4 </span>.
<span><span> -3 + 1 = -2</span><span> -1 + 3 = 2
</span></span>Final result :<span> c2 + 4c - 3 </span>
Commutative property of multiplication
So below m in the chart, is shows what m equals for you to substitute in the equation
So the second blank would be 42.
Since 7*6=42.
The third blank would equal to 56.
Since 7*8=56
The probability that none of their four children has the disease = 0.316
Step-by-step explanation:
Step 1 :
The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25
So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75
Step 2 :
There are four children. The probability that each child has the disease is independent of the condition of the other children
The probability of more than one independent events happening together can be obtained by multiplying probability of individual events
So, the probability that none of their 4 children has the sickle cell disease = 
= 0.316
Step 3 :
Answer :
The probability no children has the specified cell disease = 0.316
