Answer: 
Step-by-step explanation:
The missing figure is attached.
You can notice that the design is a cube.
A cube has 6 faces which are all squares.
The area of a square can be calculated with this formula:
Where "s" is the length of any side of the square.
In this case, you can observe in the figure that:
Therefore, substituing that value into the formula, you get that the area of any of the cube's faces is:
Finally, in order to find the surface area of the cube, you need to multiply the area of a face calculated above, by 6.
Therefore, the surface area of the design is:
Answer:
- 63 ft²
- h = 2A/b; h = 7 ft
Step-by-step explanation:
1. Put the values where the variables are in the formula, then do the arithmetic.
... A = (1/2)bh = (1/2)(18 ft)(7 ft) = 63 ft²
2. Divide by the coefficient of h. Substitute the given values and do the arithmetic.
... A = (1/2)bh
The coefficient of h is b/2, so we divide by that—or multiply by its inverse, 2/b.
... 2A/b = h
Substituting the given numbers, you have ...
... 2(28 ft²)/(8 ft) = h = 7 ft
The height of the triangle is 7 ft.
Answer:
78 and 6 tenths
Step-by-step explanation:
HOPE THIS HELPS! IF IT DOESN'T...I'M SO SORRY! SORRY 4 REALLY BAD CAMERA QUALITY! :(
HAVE A GR8 DAY ;-)
Answer:
Step-by-step explanation:
Hello!
To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.
To test this claim two samples were taken:
Sample 1
X: battery lifespan of a unit of the new product (min)
n= 93 units of the new product
mean battery life X[bar]= 8:53hs= 533min
S= 84 min
Sample 2
X: battery lifespan of a unit of the leading product (min)
n= 102 units of the leading product
mean battery life X[bar]= 5:40 hs = 340min
S= 93 min
The population variances of both variances are unknown and distinct.
To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:
H₀: μ₁ - μ₂ ≤ 120
H₁: μ₁ - μ₂ > 120
Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:
![Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7B%28X%5Bbar%5D_1-X%5Bbar%5D_2%29-%28Mu_1-Mu_2%29%7D%7B%5Csqrt%7B%5Cfrac%7BS_1%5E2%7D%7Bn_1%7D%20%2B%5Cfrac%7BS_2%5E2%7D%7Bn_2%7D%20%20%7D%20%7D)
Z≈N(0;1)
I hope this helps!
