Answer:
Last year there were 550 bison in the heard and this year it is 10% larger.
10% is also 10/100 or 0.10
To find 10% of the heard last year
550 x.0.10 = 55. To find the number in the heard this year add 550 (last year) + 55 (10% growth) = 605 or,
to find how many there are now last years heard size was 100% of the heard or 1.00. To find any percent more add it to 1.00 and multiply by the starting amount.
550 x 1.10 = 605
For the second example.
If there are 550 bison in the heard this year, and this year was a decrease of 10% from last year we can express last year's number like this.
550 = (x - (x * 0.10)) simplify;
550 = 1x - 0.10x
550 = 0.90x
Divide both sides by 0.90
x = 611.11 but it's hard to have 0.11 of a bison so realistically 611 bison last year. Subtract 10% or 61 and you indeed have 550 this year.
Answer:
208 feet
Step-by-step explanation:
We have to multiply 4x4 to get 16, and since there looks like there are 5 small squares we multiply 5x16 to get 80. And then there are 2 bigger squares that are 8x8 to get 64, then we multiply by 2 to get 128, then we add 80 (the sum of the 5 smaller squares) to get to a total of 208.
Answer:
B is
Step-by-step explanation:
Use function Y=ax, try it on B and you will get a=-1.
others are not linear function.
Answer:
A.
Step-by-step explanation:
This equation has no solution.
<h2>
Answer:</h2>
<h2>
Step-by-step explanation:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>
(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is 
</u></em>
