Answer:
7,000,000
7,800,000
5,350,000
Step-by-step explanation:
1. The function that models this equation is:
Since the value V (x) of the copier depreciates over
time, hence the 500 must be negative, therefore the equation:
V(x) = −500x + 4800
2. The rate of change for the function is:
The rate of change is also equivalent to the slope of the
equation, in this case the slope is equal to:
-$500/1 year
3. The model predicts that the value of the copier after
3 years will be:
Substituting the value of x = 3 to the function gives us:
V(x) = −500 (3) + 4800
V(x) = $3,300
4. The model predicts that the value of the copier after
7 years will be:
Substituting the value of x = 7 to the function gives us:
V(x) = −500 (7) + 4800
V(x) = $1,300
F(x) = 1600/400 reduces to 4....the slope of f(x) is 4
g(x) :
(1,-8)(5,12)
slope = (12 - (-8) / (5 - 1) = (12 + 8)/4 = 20/4 = 5...slope of g(x) is 5
so g(x) has a greater slope then f(x)
Answer:
x = 28 cm
Step-by-step explanation:
Given:
Area of link shaded regions = 84 cm²
Required:
The value of x (diameter of the semicircle/length of the rectangle)
Solution:
Diameter of the semicircle = 2r = x
Length of rectangle (L) = 2r = x
Radius of semicircle (r) = ½x
Width of rectangle (W) = radius of semicircle = ½x
Use 3.14 as π
Area of the link shaded regions = area of rectangle - area of semicircle
Thus:
Area of the link shaded regions = (L*W) - (½*πr²)
Plug in the values
84 = (x*½x) - (½*3.14*(½x)²)
84 = x²/2 - (1.57*x²/4)
84 = x²(½ - 1.57/4)
84 = x²(0.5 - 0.3925)
84 = x²(0.1075)
Divide both sides by 0.1075
84/0.1075 = x²
781.4 = x²
√781.4 = x
27.9535329 = x
x = 28 cm
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!