What is b-1 over b-1

A bag of candy contains 300 pieces and 25% of the pieces in the bag are red

taurus [48]

Answer:

75 of them are red

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the perimeter of the polygon. Round your answer to the nearest tenth.

dimulka [17.4K]

The perimeter of the polygon is 26.1 units.

To find the perimeter of the polygon.

<h3>What is polygon?</h3>

A polygon is a two-dimensional geometric figure that has a finite number of sides. The sides of a polygon are made of straight line segments connected to each other end to end. Thus, the line segments of a polygon are called sides or edges.

Given that:

Distance between points $(x_{1} ,y_{1} )$ , $(x_{2} ,y_{2} )$ is given by

$\sqrt{(x_{2} -x_{1} )^{2}+(y_{2} -y_{1} )^{2} }$

As per the graph,

point Q is (2,0)

point R is (4,5)

point S is (8,7)

point T is (6,4)

point U is (10,3)

QR= $\sqrt{(4 -2 )^{2}+(5 -0 )^{2} }=\sqrt{4+25} =\sqrt{29}$ $=5.38 units$

RS= $\sqrt{(8 -4 )^{2}+(7 -5 )^{2} }=\sqrt{16+4} =\sqrt{20}$ $=4.47 units$

ST= $\sqrt{(6 -8 )^{2}+(4 -7)^{2} }=\sqrt{4+9} =\sqrt{13}$ $=3.60 units$

TU= $\sqrt{(10 -6 )^{2}+(3 -4 )^{2} }=\sqrt{16+1} =\sqrt{17}$ $=4.12 units$

QU= $\sqrt{(10 -2 )^{2}+(3 -0 )^{2} }=\sqrt{64+9} =\sqrt{73}=8.54 units$

Perimeter refers to sum of lengths of all the sides.

Perimeter =5.38+4.47+3.60+4.12+8.54=26.1 units

So. the perimeter of the polygon is 26.1 units.

Learn more about polygon here:

brainly.com/question/28193374

#SPJ1

8 0

2 years ago

Calculate the cost of spraying a rectangular field 720 m by 500 m with a pesticide at a cost of $23.50 per hectare.

Drupady [299]

(I) $ 846

(II) 157.64 cm²

Step-by-step explanation:

(I)

The area of the rectangle is;

720 * 500 = 360,000

10,000 m2 = 1 hectare

Therefore to convert the area to hectare;

360,000 / 10,000

= 36

If the cost of spraying the pesticide is $23.50 per hectare, then total cost is;

36 * 23.50

= 846

(II)

To find the area of the trapezoid;

(sum of the parallel sides * height) / 2

Height is the diameter of the circle – 21 cm

((19 + 29)*21) / 2

= 504

Then the area of the circle;

πr²

π * (21/2)²

= 346.36

The shaded area;

Area of the trapezoid – area of the circle

504 – 346.36

= 157.64

Learn More:

brainly.com/question/10676678

brainly.com/question/12389990

#LearnWithBrainly

4 0

3 years ago

Can you please solve for b. y=bx-3

jeyben [28]

Answer: b=y+3/x

Step-by-step explanation:

bx-3=y

bx-3+3=y+3

bx=y+3

bx/x=y+3/x

6 0

3 years ago

The Highway Safety Department wants to study the driving habits of individuals. A sample of 37 cars traveling on a particular st

kodGreya [7K]

Answer:

90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

Step-by-step explanation:

We are given that a sample of 37 cars traveling on a particular stretch of highway revealed an average speed of 70.7 miles per hour with a standard deviation of 6.3 miles per hour.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average speed of cars = 70.7 miles per hour

s = sample standard deviation = 6.3 miles per hour

n = sample of cars = 37

$\mu$ = true mean speed

<em>Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

So, 90% confidence interval for the true mean, $\mu$ is ;

P(-1.688 < $t_3_6$ < 1.688) = 0.90 {As the critical value of t at 36 degree of

freedom are -1.688 & 1.688 with P = 5%}

P(-1.688 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.688) = 0.90

P( $-1.688 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.688 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.688 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.688 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<em><u>90% confidence interval for</u></em> $\mu$ = [ $\bar X-1.688 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.688 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $70.7-1.688 \times {\frac{6.3}{\sqrt{37} } }$ , $70.7+1.688 \times {\frac{6.3}{\sqrt{37} } }$ ]

= [68.9517 miles per hour , 72.4483 miles per hour]

Therefore, 90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

<em>The interpretation of the above interval is that we are 90% confident that the true mean speed of all cars will lie between 68.9517 miles per hour and 72.4483 miles per hour.</em>

8 0

3 years ago