Judy has a sugar cone and wants to know how many cubic inches of ice cream it will hold if it is filled completely to the top of the cone and no more. the cone has a height of 4.5 inches and a radius of 1.5 inches.

Transformation involves changing the position of a shape

The transformation rule is: [tex](x-10,y-3)[/tex]

From the attachment, we have:

[tex]A = (4,7)[/tex]

[tex]A' = (-6,4)[/tex]

The translation rule from ABCD to A'B'C'D' is calculated as follows:

[tex](x,y) = A' - A[/tex]

This gives:

[tex](x,y) = (-6,4) - (4,7)[/tex]

Rewrite as:

[tex](x,y) = (-6- 4,4-7)[/tex]

[tex](x,y) = (-10,-3)[/tex]

Hence, the transformation rule is:

[tex](x-10,y-3)[/tex]

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Answer:

The fraction 14/5 can be expressed as 280 percent.

Answer:

Option B is correct.    

Step-by-step explanation:

We have been given the different probabilities at different point we need to find the probability of not of 2

We know that sum of probabilities is 1

[tex]P(a_2)+P(not a_2)=1[/tex]

Here, the probability that is P(a 2)=0.40

Hence, the required probability is 1-0.40=0.60

Therefore, option B is correct.

Comparing 59 pounds to 35 pounds shows a decrease of 24 pounds, which can be calculated by subtracting the smaller number from the larger number.

If we are considering the transition from 59 pounds to 35 pounds, we need to determine whether this change represents an increase or a decrease. Comparing the two numbers, we can observe that 35 pounds is less than 59 pounds. Therefore, moving from a higher number to a lower number indicates a decrease. To calculate the decrease, we subtract the smaller number (35 pounds) from the larger number (59 pounds) which equals 24 pounds. So, there is a decrease of 24 pounds.

Using the example parameters given in the hypothetical experiment on weight perception, we've applied similar reasoning. If someone in the experiment stepped down from lifting 20 pounds to lifting weights less than this, it would also be considered a decrease in weight. If they stepped up from lifting a weight of 20 pounds to a heavier weight, it would be termed an increase. This example underscores the importance of difference detection, which is often studied in sensory experiments within psychology.

Answer:

27

Step-by-step explanation:

Convert the fraction in the exponent to a root, to simplify the expression and make the calculation easier.

[tex]9^{\frac{3}{2}}\\=\sqrt{9^{3} } \\=\sqrt{729} \\=27[/tex]

Therefore the integer which is equivalent to [tex]9^{\frac{3}{2} }[/tex] is 27.

we know that

The simple interest formula is equal to

[tex]A=P(1+rt)[/tex]

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

[tex]t=9\ years\\ P=\$2,136\\ A=?\\r=0.0536[/tex]

substitute in the formula above

[tex]A=2,136(1+0.0536*9)[/tex]

[tex]A=2,136(1.4824)[/tex]

[tex]A=\$3,166.41[/tex]  

Round to the nearest dollar

[tex]A=\$3,166[/tex]  

therefore

the answer is the option D

[tex]\$3,166[/tex]  

Answer: Experimental probability of tossing head is [tex]\frac{11}{20}[/tex]

Step-by-step explanation:

Since we have given that

Number of times Lynn tossed a coin = 60 times

Number of times head comes = 33

Experimental probability of tossing heads using Lynn and drawn results is given by

[tex]\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\\\=\frac{33}{60}\\\\=\frac{11}{20}[/tex]

Hence, Experimental probability of tossing head is [tex]\frac{11}{20}[/tex].

Answer:

experimental probability of tossing heads using Lynn and dawns results is [tex]\frac{11}{20}[/tex].

Step-by-step explanation:

Given :Lynn and dawn tossed a coin 60 times and got heads 33 times

To find : what is the experimental probability of tossing heads using Lynn and dawns results.

Solution : We have given that  

Number of times Lynn tossed a coin = 60 times.

Number of times head comes = 33.

 Probability of tossing heads using Lynn and drawn results is given by:

  = N[tex]\frac{number of favouable outcome }{total possible outcome}[/tex]

= [tex]\frac{33}{60}[/tex].

On simplification

=[tex]\frac{11}{20}[/tex].

Therefore, experimental probability of tossing heads using Lynn and dawns results is [tex]\frac{11}{20}[/tex].

Answer:

160g

Step-by-step explanation:

Answer:

Step-by-step explanation:

(-)(-) = (+)

(-)(+) = (+)(-) = (-)

(+)(+) = (+)

2 - (-8) + (-3) = (*)

-(-8) = 8

+(-3) = -3

(*) = 2 + 8 - 3 = 10 - 3 = 7

Answer:

C. 17.0 feet

Step-by-step explanation:

We have been given that Jacob kicks a soccer ball off the ground and in the air with an initial velocity of 33 feet per second. We are asked to find the maximum height the soccer ball using formula [tex]H(t)=-16t^2+vt+s[/tex].

First of all, we will substitute [tex]v=33[/tex] in our given formula.

[tex]H(t)=-16t^2+33t+0[/tex]

Since our given parabola has a negative leading coefficient, so it will be downward opening parabola. The maximum height of the ball will be y-coordinate of the vertex of parabola.

Let us find x-coordinate of parabola as:

[tex]\frac{-b}{2a}=\frac{-33}{2\times -16}=\frac{-33}{-32}=\frac{33}{32}[/tex]

Now, we will substitute [tex]x=\frac{33}{32}[/tex] in our formula to find y-coordinate of vertex.

[tex]H(\frac{33}{32})=-16(\frac{33}{32})^2+33(\frac{33}{32})+0[/tex]

[tex]H(\frac{33}{32})=-16*\frac{1089}{1024}+\frac{1089}{32}[/tex]

[tex]H(\frac{33}{32})=-16*1.0634765625+34.03125[/tex]

[tex]H(\frac{33}{32})=-17.015625+34.03125[/tex]

[tex]H(\frac{33}{32})=17.015625[/tex]

[tex]H(\frac{33}{32})\approx 17.0[/tex]

Therefore, the ball reached the maximum height of 17.0 feet and option C is the correct choice.

That makes no sense, what DO you get?

The correct answer is, A, or question 1, or 2w + 3.

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Answer: Option 'D' is correct.

Step-by-step explanation:

Since we know that

Probability is the giving the possibility of given outcome or any certain events.

Probability is the number of ways of achieving the success.

Probability of event that certain to happen is 1.

Probability of event that is impossible is 0.

Sum of probabilities of all likelihood occurring events is 1.

So, Probability of any event can exceed 1.

But,

[tex]\frac{5}{4}=1.25\text{ which is greater than 1}\\[/tex]

Hence, it is not possible value for a probability .

Therefore, Option 'D' is correct.

Final answer:

The value 5/4 is not a valid probability, as probabilities must be between 0 and 1 inclusive. The correct answer is D.

Explanation:

The question that needs assistance asks about which value is not a possible value for a probability. By definition, probabilities can range from 0 (the event will not occur) to 1 (the event is certain to occur). So, looking at the options given, A (0.001), B (1/16), and C (0.82) are all between 0 and 1, which means they are valid probabilities. However, option D (5/4) is greater than 1, which makes it an invalid probability. The correct answer is D: 5/4 is not a possible value for a probability.

An expression when written in either radical form or rational exponent form be rewritten to fit the other form as well.

When we write in different forms the Denominator defines as the Index  and the Numerator defines as Power on the variable.

  We can write  [tex]4^{\frac{2}{3}[/tex]  as   [tex]\sqrt[3]{4^2}=\sqrt[3]{16}=\sqrt[3]{8*2}=2\sqrt[3]2}[/tex]

Again, vice versa,

   We can write [tex]\sqrt[5]{x^4}[/tex] as [tex]x^{\frac{4}{5}[/tex]

Therefore , we can written in other forms as well to fit .

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To convert from radical form to rational exponent form, use [tex]\( \sqrt[n]{a} = a^{1/n} \),[/tex] and vice versa for conversion.

An expression written in radical form can be rewritten in rational exponent form and vice versa using the following conversions:

1. From Radical Form to Rational Exponent Form:

  - For a radical expression [tex]\( \sqrt[n]{a} \), where \( n \)[/tex] is the index and [tex]\( a \)[/tex] is the radicand:

  - The equivalent expression in rational exponent form is [tex]\( a^{1/n} \)[/tex].

2. From Rational Exponent Form to Radical Form:

  - For an expression [tex]\( a^{m/n} \)[/tex], where [tex]\( a \)[/tex] is the base, [tex]\( m \)[/tex] is the numerator, and [tex]\( n \)[/tex] is the denominator:

  - The equivalent expression in radical form is [tex]\( \sqrt[n]{a^m} \).[/tex]

These conversions allow us to switch between radical form and rational exponent form easily. It's important to remember that the index of the radical corresponds to the denominator of the rational exponent, and the exponent of the base corresponds to the numerator of the rational exponent.

The dimensions of the right-circular cylinder, which has a volume of 1 liter and uses the least amount of material, are such that the radius and twice the height are equal to the cube root of the volume divided by π, i.e., (0.001/π)^(1/3).

In Mathematics, given the volume of a right-circular cylinder (an open top can), we can find its optimal dimensions that would use the least amount of material. These dimensions correspond to the minimum surface area of the cylinder, which includes its closed bottom but not the open top.

Since the volume V of a right-circular cylinder is given by V=πr²h, where r is the radius and h is the height. We know that the volume equals 1 liter or 0.001 m³. Rearranging the volume equation for h gives us h=V/(πr²).

Next, the surface area A of the cylinder with closed bottom is A=2πrh+πr². Substituting h=V/(πr²) into the surface area gives A=2r(V/r)+(πr²) which simplifies to A=2V/r+πr². For the surface area to be minimum, the derivative of A with respect to r must be equal to zero. The first derivative of A is A'=-2V/r²+2πr. Set A'=0 we solve for r we find r=(V/(π))^(1/3). Substituting this value back into the equation for h gives us h=2*(V/(π))^(1/3).

Hence, for a right-circular cylinder of 1 liter volume (0.001 m³) to use the least amount of material, it should have a radius and twice the height equal to the cube root of the volume divided by π, i.e., (0.001/π)^(1/3).

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yep 20.4 is the correct answer for plato!!

we will proceed to verify each of the cases to determine the solution of the problem

we know that

The combination of raisins and chocolate chips that can the baker use, must be inside the triangular region

so

we're graphing each of the cases

Let

x--------> the number of possible raisins

y--------> the number of possible chocolate chips

case A

Let

[tex]A(28,20)[/tex]

using a graphing tool

see the attached figure

The point [tex]A(28,20)[/tex] is  include inside the triangular region

therefore

The case A is a solution

case B

Let

[tex]B(33,25)[/tex]  

using a graphing tool

see the attached figure

The point [tex]B(33,25)[/tex] is not include inside the triangular region

therefore

The case B is not a solution

case C

Let

[tex]C(45,10)[/tex]  

using a graphing tool

see the attached figure

The point [tex]C(45,10)[/tex] is not include inside the triangular region

therefore

The case C is not a solution

case D

Let

[tex]D(55,12)[/tex]    

using a graphing tool

see the attached figure

The point [tex]D(55,12)[/tex] is not include inside the triangular region

therefore

The case D is not a solution

the answer is the option A

[tex](28,20)[/tex]