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lana66690 [7]
4 years ago
14

Josh make the rectangular banner. It is 3/4 yards long and 1/4 yards wide. What is the area in square yards of the banner?

Mathematics
1 answer:
dybincka [34]4 years ago
3 0
0.1875 is the area
L*W
0.75*0.25
0.1875

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Solve for x: 3x-10 = 2(x+6)
77julia77 [94]

Answer:

x = 22

Step-by-step explanation:

7 0
3 years ago
Two sailboats had a race. Sailboat A traveled 40 feet every 5 seconds, and sailboat B traveled 56 feet every 8 seconds. Which sa
Novosadov [1.4K]

Answer:

Sailboat A won the race, and it was going 8 feet per second.

Step-by-step explanation:

Sailboat A was traveling at 40 feet per every 5 seconds, so, to find how many feet it was going per second, we would do 40 feet/5 seconds, which is 8 feet.

Sailboat B was traveling at 56 feet per every 8 seconds, so we use the same method. 56 feet/8 seconds is 7 feet.

So, Sailboat A won because it was going faster! :)

6 0
2 years ago
The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c
Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle.  We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}

\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01

3 0
4 years ago
= – 1 is a solution of the equation
Nina [5.8K]

Answer:

C

Step-by-step explanation:

substitute -1 into all of the formulas, if both sides are equal, then it is correct, for C:

2(x-2)+6 = 0, sub -1

2(-1-2)+6=0, simplify and work out

2(-3)+6=0

-6+6=0

0=0

4 0
3 years ago
According to a recent​ study, 1 in every 10 women has been a victim of domestic abuse at some point in her life. Suppose we have
Schach [20]

According to the given, 1 in every 10 women has been a victim, therefore the probability of a  victim of domestic abuse at some pint is

p=\frac{1}{10} = 0.10

We need to use Binomial distribution to take a randomly sampled of 25 women and asked each whether she has been a victim.

f(x)= ^nC_x (p)^x (1-p)^{n-x}

Where x is the event that shows a victim of domestic abuse at some point.

We need to find the probability of these victim at least 2 has been victim of abuse, so

P(x>=2) = 1- [P(x=0)+P(x=1)]

P(x>=2) = 1 - [^{25}C_0(0.10)^0(1-0.10)^{25-0}+^{25}C_1(0.10)^1(1-0.10)^{25-1}]

P(x>=2) = 1 - [0.07198+0.199]

P(x>=2) = 0.72902

So the probability is 0.729

3 0
4 years ago
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