Increasing. For every week that goes by, Olivia's puppy is gaining one pound. 6-5= 1 14-13= 1. Gaining a pound every week makes the puppies weight increase. Since increase is the term for something is bigger than number or value than before, Olivia's puppies' weight is increasing.
I hope this helps!
~kaikers
Answer:
3
Step-by-step explanation:
cause if she writes 4 posts and kevin writes 21 than she only have 3 posts left that she have write
Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: speed of a vehicle along a stretch of I-10 (mph)
This variable has a normal distribution with mean μ= 81 mph and a standard deviation σ= 8 mph.
The speed limit in the said stretch is 65 mph.
You need to calculate the probability of picking a car at random and its speed be at most 65 mph, symbolically:
P(X≤65)
To reach the probability, you need to use the standard normal distribution. To standardize the value fo X you have to subtract the value of μ and then divide it by σ:
P(Z≤(65-81)/8)= P(Z≤-2.00)
Now you look for the corresponding probability in the table of the standard normal distribution, since the value is negative you have to use the left entry. The integer and first decimal numbers are in the first column and the second decimal number is in the first row.
P(Z≤-2.00)= 0.0228
I hope it helps!
Answer: A Single Solution
Step-by-step explanation:
The question is somewhat poorly posed because the equation doesn't involve <em>θ</em> at all. I assume the author meant to use <em>x</em>.
sec(<em>x</em>) = csc(<em>x</em>)
By definition of secant and cosecant,
1/cos(<em>x</em>) = 1/sin(<em>x</em>)
Multiply both sides by sin(<em>x</em>) :
sin(<em>x</em>)/cos(<em>x</em>) = sin(<em>x</em>)/sin(<em>x</em>)
As long as sin(<em>x</em>) ≠ 0, this reduces to
sin(<em>x</em>)/cos(<em>x</em>) = 1
By definition of tangent,
tan(<em>x</em>) = 1
Solve for <em>x</em> :
<em>x</em> = arctan(1) + <em>nπ</em>
<em>x</em> = <em>π</em>/4 + <em>nπ</em>
(where <em>n</em> is any integer)
In the interval 0 ≤ <em>x</em> ≤ 2<em>π</em>, you get 2 solutions when <em>n</em> = 0 and <em>n</em> = 1 of
<em>x</em> = <em>π</em>/4 <u>or</u> <em>x</em> = 5<em>π</em>/4