1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok
2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No
3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No
4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok
5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No
Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10
Answer:
x = 12, y =
Step-by-step explanation:
a || b
5x - 7 = 3x + 17 Alternative Interior Angles
x = 12 Algebra
Angle 3x + 17 = 53 Substitution
53 + 4y + 3 = 180 Supplementary Angles
y = 31
Fatima's claim is not supported by the table because, the distribution is skewed right, with a median of 0.4 field goal advantage.
From the table, the median position is calculated as:
The 0.2nd data falls in the 0.4 field goal category.
So, the median element is:
However, the distribution of the table are concentrated on the left.
This means that, the distribution is not uniform, instead it is skewed right.
A uniform distribution has a skewness of 0.
Hence, Fatima's claim is not supported by the table
Read more about distributions at:
brainly.com/question/13233983
Answer: The first answer
A coin bank can hold 710 coins. There are 50 coins in the bank. A student puts in 100 coins per week. How many weeks until the bank is full?
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
Labor productivity is is measured per hours.
Hence, the work he does [amount of claims] will be distributed amongst the amount of hours he works. The amount of dollar he uses <em>doesn't matter for labor productivity.</em>
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Thus, his work is 6 claims. He does that is 8 hours. So claim/hr is:
6/8 = 0.75 claims per hour
Answer choice D is right.
