The area calculation for the shaded section, as seen in the attached diagram, involves subtracting the area of the kite from that of the rectangle. The area of the rectangle is calculated as (3x+x)*(x+x), which simplifies to 4x*2x, equating to 8x². The area of the kite is determined using the formula (1/2)*[d1*d2], where d1 and d2 represent the diagonals, specifically d1=4x and d2=2x. Therefore, the area of the kite becomes (1/2)*[4x*2x], leading to 4x². Consequently, the area of the shaded region can be computed as 8x²-4x², resulting in 4x². Thus, the solution is 4x².
Malcolm's remaining distance is roughly
1370 - 470 - 430 = 470
This coincides with selection...
D) 470 miles
The diagonal measures 20.68 ft; the shorter base is 17.21 ft. To understand this, we recognize that with base angles summing to 140°, each angle is 70°, given the isosceles trapezoid's properties. We can apply the Law of Cosines to find the diagonal's length, denoted as d. The length of the diagonal determines to be d = 20.68 ft. Determining the shorter base is somewhat more complex. By drawing an altitude from the upper vertices to the base, which measures 22 ft, we create two similar smaller right triangles requiring us to find the height and base measures related to each of the 70-degree angles and the hypotenuse of 7. By working through the calculations for height and base from one triangle, we subsequently find that 22 minus twice the base measure gets us to the shorter base's measure, arriving at x = 17.21 ft.
Each LED bulb, along with installation labor, is priced at
.. $6.95 +$3 = $9.95
For 100 bulbs over a span of 10 years, that equals (100*10) = 1000 bulb·years. At $9.95 per bulb, 5 bulb·years are obtained, and thus the projected total cost for 1000 bulb·years is
.. (1000 b·y)*($9.95/(5 b·y)) = $1990
In summary, for a decade, the installation and changes of 200 bulbs in 100 lamps amount to $1990. Therefore, the yearly cost is...
.. $1990/(10 yr) = $199/yr
Answer:
The number of standard deviations above the mean is 
Step-by-step explanation:
The question indicates that:
The average weight of the corn ears from each farm is 
The standard deviation for the corn ears from Iowa is 
The standard deviation for the corn ears from Ohio is
A randomly chosen ear of corn from Iowa weighs x = 1.39 pounds
The standardized score is z = 1.645
The weight of a randomly chosen ear of corn from Ohio measures 
In general, the standardized score of corn weight from Iowa can be mathematically defined as:
=> 
=> 
=> 
Conversely, the standardized score of corn weight from Ohio is expressed as:
=> 
=> 
=>
A positive value indicates this quantity represents the number of standard deviations above the mean.
