Answer:
Answer:
A) Total frequency = 30
B) Yes, the distribution is consistent with the label because the frequencies are greatest when the lengths are closest to the labeled size of 1/2 inches which is 0.5 inches.
Step-by-step explanation:
Since the class limit width is 0.010 from the question, we arrive at;
Class Limits Frequency
Length(Inches)
0.470 - 0.479 3
0.480 - 0. 489 5
0.490 - 0.499 9
0.500 - 0.509 12
0.510 - 0.519 1
Answer:
Y=10x
(multiply 10 times the salamanders)
Step-by-step explanation:
a formula that can be used for this circumstance is a linear y=mx+b problem. for 1 salamander there are 10 frogs. so with this we can state that y=10x. whatever you plug into the x for the number of salamanders will give you the answer of how many frogs are in the pond. for example.
Y=10x
Y=10(3)
Y=30
30 frogs.
also because 10 is a nice number you can just multiply the number of salamanders times 10. 3*10=30
Look up desmos, it’ll graph it :)
Answer:
log₁.₅3 or 2.70951
Step-by-step explanation:
- 3^(2x) - 2^(x + 1) * 3^x - 3*2^(2x) = 0
- 3^(2x)- 2*2^x*3^x + 2^(2x)- 4*2^(2x)=0
- (3^x-2^x)^2 - (2*2^x)^2=0
- (3^x-2^x+2*2^x)(3^x - 2^x - 2*2^x)=0
- (3^x+2^x)(3^x- 3*2^x)=0
- 3^x+2^x>0 for any value of x
- 3^x- 3*2^x=0
- 3^x= 3*2^x
- 3^x/2^x=3
- 1.5^x=3
- x= log₁.₅ 3 = 2.70951