Answer:
25
Step-by-step explanation:
had it on khan
Let us see... ideally we would like to have all equations with the same exponent or the same base so that we can compare the rates. Since the unknown is in the exponent, we have to work with them. In general,
![x^(y/z)= \sqrt[z]{x^y}](https://tex.z-dn.net/?f=x%5E%28y%2Fz%29%3D%20%5Csqrt%5Bz%5D%7Bx%5Ey%7D%20)
.
Applying this to the exponential parts of the functions, we have that the first equation is equal to:
250*(
![\sqrt[5]{1.45} ^t](https://tex.z-dn.net/?f=%20%5Csqrt%5B5%5D%7B1.45%7D%20%5Et)
)=250*(1.077)^t
The second equation is equal to: 200* (1.064)^t in a similar way.
We have that the base of the first equation is higher, thus the rate of growth is faster in the first case; Choice B is correct.
Answer:
Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.
A. The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.
C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.
E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake
The answer is E.
Answer:
Option (C)
Step-by-step explanation:
Value of y is more than the product of x and 2.
Equation representing the given condition will be,
y = 2x + 1
Therefore, by substituting the values of x in the equation we can get the table for the input - output values for the equation,
x 0 1 2 3 4
y 1 3 5 7 9
Therefore, table given in the Option (C) will be the correct option.
NOTE THIS IS AN EXAMPLE:
Let t = time, s = ostrich, and g = giraffe.
Here's what we know:
s = g + 5 (an ostrich is 5 mph faster than a giraffe)
st = 7 (in a certain amount of time, an ostrich runs 7 miles)
gt = 6 (in the same time, a giraffe runs 6 miles)
We have a value for s, so plug it into the first equation:
(g + 5)t = 7
gt = 6
Isolate g so that we can plug that variable value into the equation:
g = 6/t
so that gives us:
(6/t + 5)t = 7
Distribute:
6 + 5t = 7
Subtract 6:
5t = 1
Divide by 5:
t = 1/5 of an hour (or 12 minutes)
Now that we have a value for time, we can plug them into our equations:
1/5 g = 6
multiply by 5:
g = 30 mph
s = 30 + 5
s = 35 mph
Check by imputing into the second equation:
st = 7
35 * 1/5 = 7
7 = 7
