Answer: First option.
Step-by-step explanation:
The complete exercise is attached.
In order to solve this exercise, it is necessary to remember the following property:
The Multiplication property of Equality states that:
In this case, the equation that Jada had is the folllowing:
Jada needed to solve for the variable "x" in order to find its value.
The correct procedure to solve for for "x" is to multiply both sides of the equation by 108. Then, you get:
As you can notice in the picture, Jada did not multiply both sides of the equation by 108, but multiplied the left side by 
<em> </em>and the right side by 
.
Therefore,you can conclude that Jada should have multiplied both sides of the equation by 108.
Answer : y>0
f(x) = 9*2^x
f(x) is an exponential function
When we plug in positive value for x , the value of y is positive
When we plug in negative value for x , the value y is also positive
So for any value of x, the y value is positive always.
Range is the set of y values for which the function is defined
y values are positive , so range is y >0
Incomple question. However, here's the remaining part of the question:
14
2009
Meadow Fritillary= 5
Variegated Fritillary= 7
Zebra Swallowtail= 33
Eastern-Tailed Blue= 242
Louden County Butterfly Count
2010
Meadow Fritillary 34
Variegated Fritillary 95
Zebra Swallowtail 21
Eastern-Tailed Blue 168
2011
Meadow Fritillary
Variegated Fritillary
Zebra Swallowtail
Eastern-Tailed Blue
10
170
<u>Options</u>:
A) All butterfly populations are steadily decreasing.
B)All butterfly populations were larger than usual in 2010.
C)The Eastern-Tailed Blue butterfly is more common than the others.
D)The Meadow Fritillary is equally common as the Variegated Fritillary
Answer:
<u>C</u>
Step-by-step explanation:
Looking through the above count data by Louden County Wildlife Conservancy from 2009 to 2011 we notice the Eastern- Trailed Blue butterfly has a higher count, which implies that the Eastern-Tailed Blue butterfly is more common than the other butterflies.
Therefore, we could infer from the samples, that the Eastern-Tailed Blue butterfly is more common than others from the records of the past 3 three years.
<span>In a baseball season:
Peter hit = 2 (x - 6)
Total = 19 hits
X= Alex’s hit
Y = Peter’s Hit
=> Y = 2(x - 6)
=> x + y = 18
Let’s start from the second given equation:
=> x + y =18
=> y = 18 – x
Now, let’s use the first given equation to solve:
=> 18 – x = 2 (x - 6)
=> 18 – x = 2x – 12
=> -3x = -30
x = 10
Now, Let’s try:
=> Y = 2(x - 6)
=> y = 2 (10 - 6)
=> y = 2 (4)
=> y = 8
Thus, Peter hit 8 and Alice hit 10.</span>
Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is 
As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is 
. Thus, there are only 35 ways to distribute the blackboards in this case.
