Answer:
<h2>7</h2>
Step-by-step explanation:
![\left[\left(11\:-\:4\right)^3\right]^2\:\div \left(4\:+\:3\right)^5\\\\\frac{\left(\left(11-4\right)^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Subtract\:the\:numbers:}\:11-4=7\\\\=\frac{\left(7^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Add\:the\:numbers:}\:4+3=7\\\\=\frac{\left(7^3\right)^2}{7^5}\\\\\left(7^3\right)^2=7^6\\\\=\frac{7^6}{7^5}\\\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\frac{7^6}{7^5}=7^{6-5}\\\\\mathrm{Subtract\:the\:numbers:}\:6-5=1\\\\=7](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft%2811%5C%3A-%5C%3A4%5Cright%29%5E3%5Cright%5D%5E2%5C%3A%5Cdiv%20%5Cleft%284%5C%3A%2B%5C%3A3%5Cright%29%5E5%5C%5C%5C%5C%5Cfrac%7B%5Cleft%28%5Cleft%2811-4%5Cright%29%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A11-4%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BAdd%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A4%2B3%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cleft%287%5E3%5Cright%29%5E2%3D7%5E6%5C%5C%5C%5C%3D%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%7D%3A%5Cquad%20%5Cfrac%7Bx%5Ea%7D%7Bx%5Eb%7D%3Dx%5E%7Ba-b%7D%5C%5C%5C%5C%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%3D7%5E%7B6-5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A6-5%3D1%5C%5C%5C%5C%3D7)
Answer:
A - 0%
B- 50%
C- 50%
D- 100%
Step-by-step explanation:
Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.
Performing the cross of Aa x AA, we can see that:
a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.
b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.
c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.
d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.
Answer:
Hey there, Lets solve this step by step.
A number t divided by 82
Let the number be represented by y.
t divided by 82 = t / 8
Therefore-
In algebraic expressions, "y" is normally used as the dependent variable and "x" is normally used as the independent variable, instead of other variables. So that is why i used the variable y in the equation.
Step-by-step explanation:
Let x be the number of households with TV.
So .68x would be the number of households with cable.
If 8000 were added to the above number, it would be equal to 67,600,000.
Thus we have our equation:
.68x + 8000 = 67600000
Solving for x:
.68x = 67592000
x = 99,400,000
