A Stats test has 5 multiple choice questions with four choices with one correct answer each. If we just randomly guess on each of the 5 questions, what is the probability that you get exactly 2 questions correct

Answer:

1,130,400 people

Step-by-step explanation:

Find the area of the broadcasting area, using the circle area formula.

Since the diameter is 120 miles, the radius will be 60 miles.

Use the circle area formula, plug in 60 as r, and plug in 3.14 as pi:

A = [tex]\pi[/tex]r²

A = (3.14)(60²)

A = 11304

So, the area is 11304 square miles

To find how many people are in receiving distance, multiply this by 100, since there are 100 people per square mile.

11304(100)

= 1,130,400

So, 1,130,400 people are within receiving distance of the signal.

Answer:

1,130,400

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

Answer:

the second store has the better deal.

Step-by-step explanation:

12 ÷ 2 = 6

7.5 × 6 = 45

20.2 < 45

Answer:

below

Step-by-step explanation:

basically find the line that connects JK and KL, and since its 2/3 and -2/3 which are not opposite reciprocals, the points dont form a right triangle

Answer:

I think ir is 2 -2

Step-by-step explanation:

hope it helps

Answer:

36

Step-by-step explanation:

So to find the last angle first. The formula is a power 2+b power 2=c power 2

So 9 power 2 is 81 and 12 power 2 is 144

144+81=225

the square root of 225 is 15. So x is 15

No we just have to do 9+12+15=36.

36 is the perimeter

Answer:

The second option, external conflict w/other citizens who don't want the Cinema.

Step-by-step explanation:

The Answer is A. Hexagonal prism

God bless u! ❤️

The base is a hexagon

The top is also hexagon

The sides are rectangle

Hence the polygon is hexagonal prism

A right angled triangle with its side length is given. The sides of the triangle are 12ft , 13ft and 5ft. And we need to find out the value of cot A .

Here :-

We know :-

[tex]\tt\implies cot \theta = base / perpendicular [/tex]

Substitute the respective values :-

[tex]\tt\implies cot \theta = base / perpendicular\\\\\tt\implies\boxed{\red{\tt cot A = 12/5 }}[/tex]

Hence the required answer is 12/5.

Answer:

36 times

Step-by-step explanation:

Given

[tex]Red = 4[/tex]

[tex]Total = 7[/tex]

[tex]n = 63[/tex]

See attachment

Required

Predicted number of times to land on red

First, we calculate the probability of landing on red

[tex]Pr = \frac{Red}{Total}[/tex]

[tex]Pr = \frac{4}{7}[/tex]

The predicted number of red is the =n calculated as:

[tex]Red = Pr * n[/tex]

[tex]Red = \frac{4}{7}*63[/tex]

[tex]Red = 4 * 9[/tex]

[tex]Red = 36[/tex]

Answer:

π2,√80,12

Step-by-step explanation:

π2= 6.28

√80= about 8

12=12

Hope I could help!

Answer:

Mean = 4

Step-by-step explanation:

[tex]Mean = \frac{\sum x}{n} = \frac{2 + 5+ 5}{3} = \frac{12}{3} = 4[/tex]

Answer:

4

Step-by-step explanation:

Mean is average, so you would add every number up, then divide by the numbers you have, So 2+5+5 divided by 3 would be 4

Answer:

I'm sorry, I can help you with only this one :( ∠F = 44°

Step-by-step explanation:

∠F = 180° - (46° + 90°) = 44°

Answer:

H

Step-by-step explanation:

For F  - Median For Class 1 is 52.5 Median for Class 2 is 75 ( False )

For G - Range for class 2 is 75 Range for class 1 is 45.9 ( False )

For H - IQR for class 1 is 30.45 IQR for class 2 is 60 ( True )

For J - Minimum for Class 2 is 30 Minimum for Class 1 is 30 ( False )

Answer:

25cm

Step-by-step explanation:

one the possibility 36 - 11 =25 the answer is 25

Answer:

144 m2

Step-by-step explanation:

The bisectors of the perpendicularity of the vectors

Answer:

40 because 40 percent of 100 is 40. You can just multiply the number by the percentage $100x40%

The interquartile range(IQR) of the recorded weight of ken is calculated as 17 pounds.

The interquartile range(IQR) = third quartile(Q3) - first quartile(Q1)

The third quartile of in the box plot given is : 56 pounds

The third quartile of the data as represented in the box plot given :

39 pounds

The interquartile range(IQR) of ken's data = 56 - 39

= 17 pounds

Hence, The interquartile range(IQR) of the recorded weight of ken is calculated as 17 pounds.

Find more information about the interquartile range here

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Answer:

3    2/21

Step-by-step explanation:

used a calculator

Answer:

Life forms change over time

Step-by-step explanation:

Answer:

$70

Step-by-step explanation:

You want 175% of $40.

175% as a decimal = 1.75

1.75 times 40 = 70

Therefore Liam put $70 in savings in April

Answer:

70

Step-by-step explanation:

Multiply 40$ by 175% :)

Answer:

the divisors of 63 are

Step-by-step explanation:

1,3,7,9,21,63

Answer:

Fraction Form: 2/6 or 1/3

or Decimal Form: .333

or Percent Form: 33.333?%

Step-by-step explanation:

The divisors of 63 that are 1-6 is 1 and 3

So there are TWO numbers out of SIX.

We would turn that into the fraction 2/6 and then simplify (optional) it to 1/3.

Then we use 2/6 to convert it to multiple forms if needed.

Answer: what’s the answer ?

Step-by-step explanation:

If each cylinder is 4 feet tall and has a diameter of 1 foot then the amount of sand needed to fill the cylinders is 19 cubic feet.

Given that there are 6 cylindrical posts protecting the equipment

To find tge amount of sand needed for cylinders,

We need to find the volume of each cylinder.

The cylinder is 4 feet tall and has a diameter of 1 foot,

Height is 4 feet

Diameter is 1 foot and radius is 1/2 feet.

Volume of cylinder = πr²h

=3.14×(1/4)(4)

=3.142 cubic feet.

As there are six cylinders.

Multiply 6 with 3.142

Amount of sand needed = 6×3.142

= 18.852 cubic feet.

To learn more on Three dimensional figure click:

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Answer:

41.8°

Step-by-step explanation:

We solve using the Trigonometric function of Sine

Hypotenuse = 15cm

x = θ

Opposite = 10cm

sin θ = Opposite/Hypotenuse

sin θ = 10/15

θ = arc sin ( 10/15)

θ = 41.810314896°

θ = Approximately to 1 decimal place = 41.8°

Therefore, x = θ = 41.8°

The volume of a pyramid with a square base is 17.6 cubic centimeter.

Given the following data;

To find the volume of a pyramid with a square base;

Mathematically, the volume of a pyramid with a square base is given by the formula;

[tex]V = A(\frac{h}{3})[/tex]

Where;

First of all, we would determine the length of sides of the square base from its perimeter;

[tex]Perimeter \ of \ square = 4s\\\\9.9 = 4s\\\\s = \frac{9.9}{4}[/tex]

s = 2.475 cm

Next, we would determine the area of the square;

[tex]Area \ of \ a\ square = s^{2} \\\\Area \ of \ a\ square = 2.475^{2}[/tex]

Area of a square = 6.126 cm

Now, we would use the above value to find the volume of a pyramid with a square base.

Substituting the values into the formula, we have;

[tex]V = 6.126*(\frac{8.6}{3})\\\\V = 6.126*2.867[/tex]

Volume, V = 17.6 cubic centimeter

Therefore, the volume of a pyramid with a square base is 17.6 cubic centimeter

Find more information here: https://brainly.com/question/16627760

Answer:

Numerator :

[tex]8x^2 - 10x + 3 = 8x^2 -6x -4x + 3[/tex]

                     [tex]= 2x(4x -3)-1(4x -3) \\= (2x-1)(4x-3)[/tex]

Denominator :

[tex]6x^2 + 3x -3 = 6x^2 +6x -3x - 3[/tex]

                   [tex]=6x(x+1)-3(x+1)\\=(6x-3)(x+1)\\=3(2x-1)(x+1)[/tex]

[tex]\frac{8x^2-10x+3}{6x^2+3x -3} = \frac{(2x-1)(4x-3)}{3(2x-1)(x+1)} = \frac{(4x-3)}{3(x+1)}[/tex]

[tex]\huge\text{Hey there!}[/tex]

[tex]\large\textsf{2h - 3 }\mathsf{>}\large\textsf{ 15}[/tex]

[tex]\large\text{ADD 3 to BOTH SIDES}[/tex]

[tex]\large\textsf{2h - 3 + 3 }>\large\textsf{ 15 + 3}[/tex]

[tex]\large\text{CANCEL out: \textsf{-3 + 3} because that gives you 0}[/tex]

[tex]\large\text{KEEP: \textsf{15 + 3} because it helps you solve for your h-value}[/tex]

[tex]\large\textsf{15 + 3 = \boxed{\large\textsf{\bf 18}}}[/tex]

[tex]\large\text{NEW EQUATION: \textsf{2h} }\mathsf{>}\large\textsf{ 18}[/tex]

[tex]\large\text{DIVIDE 2 to BOTH SIDES}[/tex]

[tex]\mathsf{\dfrac{2h}{2}>\dfrac{18}{2}}[/tex]

[tex]\large\text{CANCEL out: }\mathsf{\dfrac{2}{2}}\large\text{ because it gives you 1}[/tex]

[tex]\large\text{KEEP: }\mathsf{\dfrac{18}{2}}\large\text{ because it gives you your h-value is being compared to}[/tex]

[tex]\large\text{NEW EQUATION: }\mathsf{h > \dfrac{18}{2}}[/tex]

[tex]\mathsf{\dfrac{18}{2}}[/tex]

[tex]\mathsf{= 18\div 2}[/tex]

[tex]=\boxed{\mathsf{\large\textsf {\bf 9}}}[/tex]

[tex]\boxed{\boxed{\large\textsf{Answer: }\mathsf{\bf h > 9}}}\huge\checkmark\\\\\boxed{\boxed{\rm \bf It\ it\ an\ O P E N E D\ circle\ shaded}}\\\boxed{\boxed{\rm \bf to\ the\ right\ starting\ at\ point\ 9}}\checkmark[/tex]

[tex]\large\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

[tex] \quad \quad \quad \quad\tt{2h - 3 > 15}[/tex]

[tex] \quad \quad \quad \quad\tt{2h - 3 > 15}[/tex]

[tex] \quad \quad \quad \quad\tt{2h - 3 + 3 > 15 + 3}[/tex]

[tex]\quad \quad \quad \quad\tt{2h \: \: \cancel{ \color{red}- 3 + 3} \: > 15 + 3}[/tex]

[tex]\quad \quad \quad \quad\tt{2h > 18}[/tex]

[tex]\quad \quad \quad \quad\tt{ \frac{2h}{2} > \frac{18}{2} }[/tex]

[tex]\quad \quad \quad \quad\tt{ \frac{ \cancel{ \color{red}2}h}{ \cancel{ \color{red}2}} > \frac{18}{2} }[/tex]

[tex]\quad \quad \quad \quad\tt{ h > \frac{18}{2} }[/tex]

[tex]\quad \quad \quad \quad\tt{ h > 9}[/tex]

[tex]\quad \quad \quad \quad \boxed{\tt { \color{green}h > 9}}[/tex]

________

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Answer: look at the picture

Step-by-step explanation: Hope this help :D