The linear form of that equation is y=1/4x+2. Your y-intercept should be 2 and the slope is 1/4.
Answer:
31.25 k
Step-by-step explanation:
80 % * x = 25 k
.80 x = 25 k
x = 25 k / .80 = 31.25 k
Answer:
0, for q ≠ 0 and q ≠ 1
Step-by-step explanation:
Assuming q ≠ 0, you want to find the value of x such that ...
q^x = 1
This is solved using logarithms.
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x·log(q) = log(1) = 0
The zero product rule tells us this will have two solutions:
x = 0
log(q) = 0 ⇒ q = 1
If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to <em>any</em> power.
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<em>Additional comment</em>
The applicable rule of logarithms is ...
log(a^b) = b·log(a)
Linear regression line y=2.1x+130 predicts sales based on the money spent on advertising.
Linear regression represents the relationship between two variables. the value of y depends on the value of x.
x represents the dollars spent in advertising and y represents the company sales in dollars.
We need to find out sales y when $150 spends on advertising.
Plug in 150 for x and find out y
y = 2.1 x + 130
y = 2.1 (150) + 130
y= 445
The company expects $445 in sales
The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let \displaystyle PP be the student population and \displaystyle nn be the number of years after 2013. Using the explicit formula for a geometric sequence we get
{P}_{n} =284\cdot {1.04}^{n}P
n
=284⋅1.04
n
We can find the number of years since 2013 by subtracting.
\displaystyle 2020 - 2013=72020−2013=7
We are looking for the population after 7 years. We can substitute 7 for \displaystyle nn to estimate the population in 2020.
\displaystyle {P}_{7}=284\cdot {1.04}^{7}\approx 374P
7
=284⋅1.04
7
≈374
The student population will be about 374 in 2020.
