– (8q+7)= 3 (16 — q) ?

Mathematics

2 answers:

emmasim [6.3K]3 years ago

8 0

Q = -11. Hope this helps

Ivanshal [37]3 years ago

6 0

Answer:

-11

Step-by-step explanation:

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What is the area of this figure?

nalin [4]

Answer:

290.5 square miles

Step-by-step explanation:

Consider splitting this figure into two mini rectangles and a triangle, each of given lengths;

$Dimensions of Rectangle 1 - Height ; 5 mi, Length ; 7 mi,\\Area of Rectangle 1 - Height * Length = 5 mi * 7 mi = 35 mi^2\\\\Dimensions of Rectangle 2 - Height ; 18 mi, Length ; 8 mi,\\Area of Rectangle 2 - Height * Length = 23 mi * 8 mi = 184 mi^2\\\\Dimensions of Mini Right Triangle - Height ; 11 mi, Base ; 8 + 5 = 13 mi,\\Area of Mini Right Triangle - 1 / 2 * Base * Height = 1 / 2 * 13 mi * 11 mi = 71.5 mi^2,\\\\Area of Figure = 35 mi^2 + 184 mi^2 + 71.5 mi^2 = 290.5 square miles$

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For a certain candy, 20% of the pieces are yellow, 15% are red, 20% are blue, 20% are green, and the rest are brown. a

OLEGan [10]

Answer:

Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

Brown: $\frac{25}{100}$
Yellow or Blue: $\frac{20}{100} +\frac{20}{100} = \frac{40}{100}$ ....add the numerators

Not Green: $\frac{80}{100}$ .... since the sum of all the rest is 80%

Stiped: $\frac{25}{100}$ .... there are 0 striped candies.

Assuming the ratios/percentages of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.