Make an inequality representing both constraints.
0.5x+y<20
x+y≥24
Option I: 10 candy bars for $6.75
<span>6.75/10 = 0.68 </span>
<span>Option II: 12 candy bars for $7.25 </span>
<span>7.25 / 12 = 0.60 </span>
draw an equilateral triangle that has a side length of all 4cm
Answer:
H0: μ = 5 versus Ha: μ < 5.
Step-by-step explanation:
Given:
μ = true average radioactivity level(picocuries per liter)
5 pCi/L = dividing line between safe and unsafe water
The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.
A type I error, is an error where the null hypothesis, H0 is rejected when it is true.
We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.
Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.
Answer:
see below
Step-by-step explanation:
7. f(x) = mx + b; b acts as the vertical shift moving the graph up if b is positive and down if b is negative.
f(x) = x-2 has shifted the graph of f(x) down 2 units.
g(x) = x + 7 has shifted the graph up 7 units.
If f(x) is changed to g(x) the vertical shift has gone from -2 to + 7 or (7-(-2) = 9 units up
8. f(x) = mx stretches the graph by a factor of m if m > 0 and compresses the graph by a factor of m if 0<m<1
f(x) = 3x stretches the graph by a factor of 3.
g(x) = 1/4x compresses the graph by a factor of 1/4
