A survey conducted by a song rating service finds that the percentages of listeners familiar with poker face by lady gaga who would give the song ratings of 5, 4, 3, 2, and 1 are, respectively, 50 percent, 18 percent, 19 percent, 8 percent, and 5 percent. assign the numerical values 1, 2, 3, 4, and 5 to the (qualitative) ratings 1, 2, 3, 4, and 5 and find an estimate of the probability distribution of x = this song’s rating by a randomly selected listener who is familiar with the song.

Answer:

The value of b nearest whole number is, 13.

Step-by-step explanation:

We know an [tex]\angle B=48^{\circ}[/tex] and the side adjacent to it i.e, a=12.\

In a right triangle BCA ,

the tangent(tan) of an angle is the length of the opposite side divided by the length of the adjacent side.  

i.e, [tex]\tan B=\frac{opposite}{Adjacent}=\frac{b}{a}[/tex]

Substitute the value of a=12 and [tex]\angle B=48^{\circ}[/tex] to solve for b in above expression:

[tex]\tan 48^{\circ}=\frac{b}{12}[/tex]

we have the value of [tex]\tan 48^{\circ}=1.110613[/tex]

then, [tex]1.20012724=\frac{b}{12}[/tex]

On simplify we get,

[tex]b=1.110613 \times 12=13.327356[/tex]

Therefore, the value of b nearest whole is, 13

The rate of petrol per liter is 11.2 liter

The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

Given:

In 25l car travels = 280km

For finding rate of petrol per liter we have to divide as

Rate of petrol = 280/25

Rate of petrol = 11.2 liter

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The rate of petrol usage for the car is 11.2 kilometers per litre, calculated by dividing the distance travelled, 280km, by the amount of petrol used, 25L.

The question asks to find the rate of petrol usage in kilometers per litre for a car that uses 25L of petrol to travel 280km. To calculate the fuel efficiency, you divide the distance travelled by the amount of petrol used.

Step-by-step calculation:

Distance travelled = 280 km

Amount of petrol used = 25L

Fuel efficiency (km/L) = Distance travelled / Amount of petrol used

Fuel efficiency = 280 km / 25L = 11.2 km/L

Therefore, the car's fuel efficiency is 11.2 kilometers per litre.

Final answer:

The equation relating the total amount of money S in Carlos's savings to the number of weeks W he adds money is S = 750 + 40W. By substituting W with 11, we find that after 11 weeks, Carlos will have $1,190 in his savings account.

Explanation:

The question involves creating a linear equation to represent the relationship between the total amount of money S in Carlos's savings account and the number of weeks W he has been adding money. The equation can be written as S = 750 + 40W, where 750 represents the initial amount and 40 is the amount added each week. To find the total amount of money after 11 weeks, substitute W with 11:

S = 750 + 40(11) = 750 + 440 = 1190

Therefore, after 11 weeks, the total amount of money in the savings account would be $1,190.

the answer is A

700 x 5 = 3500

30 x 5 = 150

7 x 5 = 35

3500 + 150 +35 = 3685

700 x 200 = 140,000

30 x 200 = 6000

7 x 200 = 1400

140000 + 6000 + 1400 = 147400

Using the normal distribution table, which shows the percentage of the areas to the left a normal distribution, the area to the left z = - 2.15 and area to the right of z = 1.62 are 0.0158 and 0.0526 respectively.

1.)

The area under the normal distribution curve to the left of z = 2.15 can be expressed thus :

P(Z ≤ -2.15)

Using a normal distribution table ; the area to the left is

P(Z ≤ -2.15) = 0.0158

2.)

The area under the normal distribution curve to the right of z = 1.62 can be expressed thus :

P(Z ≥ 1.62) = 1 - P(Z ≤ 1.62)

Using a normal distribution table ; the area to the left is P(Z ≤ 1.62) = 0.94738

P(Z ≥ 1.62) = 1 - 0.94738 = 0.0526

Therefore, the area to the left z = - 2.15 and area to the right of z = 1.62 are 0.0158 and 0.0526 respectively.

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Answer:

Step-by-step explanation:

The given function is

[tex]f(x)=x^{4}-32x^{2} -144[/tex]

The given zero to this function is

[tex]x=-2i[/tex]

Remember that a function has so many roots as its grade dictates. So, in this case, the function grade is four, which means it has four solutions.

Now, the given function is an imaginary number, which happens in pairs, that means the second root must be [tex]x=2i[/tex], because a function can have complex roots as pairs, 2, 4, 6,... many solutions.

The other two solutions are 6 and -6, because if we replace them into the function, it will give zero.

[tex]f(6)=6^{4}-32(6)^{2} -144=0\\f(-6)=(-6)^{4}-32(-6)^{2} -144=0[/tex]

Therefore, the other three solutions missing are: c) 2, 6, -6.

(The image attached shows the real solutions).

The cost of one Sociology workbook is $140, and the cost of one Economy 101 textbook is $220, determined by solving two simultaneous equations based on the initial and additional orders placed by the bookstore.

To solve for the cost of one Sociology workbook and one Economy 101 textbook, let's label the cost of one Sociology workbook as S and the cost of one Economy 101 textbook as E. Initially, the bookstore ordered 15 Sociology workbooks and 10 Economy 101 textbooks, spending a total of $4300. This gives us the equation: 15S + 10E = 4300. Later, the store ordered an additional 1 Sociology workbook and 3 Economy 101 textbooks for $800, leading to the equation: 1S + 3E = 800.

To solve these equations simultaneously, we can multiply the second equation by 15 to get 15S + 45E = 12000 and subtract the first equation from it: (15S + 45E) - (15S + 10E) = 12000 - 4300, simplifying to 35E = 7700. Dividing both sides by 35 gives us E = 220. Substituting E = 220 back into the first equation 15S + 10(220) = 4300 gives us 15S + 2200 = 4300, thus 15S = 2100 and S = 140.

Therefore, the cost of one Sociology workbook is $140 and the cost of one Economy 101 textbook is $220.

Answer:

16 50 min

Step-by-step explanation:

The value of d, when the function is at its minimum is -4.

A function is a law that relates two variables namely, a dependent and an independent variable.

A function always has a defined range and domain, domain is all the value a function can have as an input and range is all the value that a function can have.

The function is of various types, logarithmic, exponential, quadratic, radical etc.

The function is y = cos [tex]\rm \theta[/tex] + d

The minimum value of the function, y = cos [tex]\rm \theta[/tex] + d is -5.

The minimum value is the lowest value of the function.

The minimum value of cos [tex]\rm \theta[/tex] is at 180 degree equals to  -1

Substituting the value in the equation.

-5 = - 1 + d

d = -5 +1

d = -4

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Answer:

The expression is Price = 36 +22*d where d is each day that pass.

Step-by-step explanation:

The inicial 36 are the independant term since it's not related to how many days the car is rented. 22 is the slope since it depends on the days the car is rented.

The domain of y = sec(x) is all real numbers except for values where the cosine function is zero.

The domain of y = sec(x) is all real numbers except for values where the cosine function is equal to zero. Since the secant function is the reciprocal of cosine, it is undefined when cosine is zero.

So, the domain of y = sec(x) is all real numbers except for values of x where cos(x) = 0.

For example, if we consider the unit circle, the cosine function is equal to zero at π/2 and 3π/2. Therefore, the domain of y = sec(x) would exclude these values.

The domain of [tex]\( y = \sec(x) \)[/tex] is all real numbers except [tex]\( x = \frac{\pi}{2} + \pi \cdot n \)[/tex], [tex]\( n \)[/tex] integer.

Let's find the domain of [tex]\( y = \sec(x) \)[/tex].

The secant function, [tex]\( \sec(x) \)[/tex], is defined as the reciprocal of the cosine function, [tex]\( \cos(x) \)[/tex]. So, [tex]\( \sec(x) = \frac{1}{\cos(x)} \).[/tex]

The cosine function is defined for all real numbers, except where the denominator becomes zero, because division by zero is undefined.

So, to find the domain of [tex]\( \sec(x) \)[/tex], we need to find where [tex]\( \cos(x) \)[/tex] is not zero.

The cosine function has a range between -1 and 1. It equals zero at [tex]\( x = \frac{\pi}{2} + \pi \cdot n \)[/tex] and [tex]\( x = -\frac{\pi}{2} + \pi \cdot n \)[/tex], where [tex]\( n \)[/tex] is an integer.

So, the domain of [tex]\( \sec(x) \)[/tex] is all real numbers except where [tex]\( \cos(x) = 0 \)[/tex]. Therefore, the domain of [tex]\( \sec(x) \)[/tex] is [tex]\( x \neq \frac{\pi}{2} + \pi \cdot n \)[/tex] and [tex]\( x \neq -\frac{\pi}{2} + \pi \cdot n \)[/tex], where [tex]\( n \)[/tex] is an integer.

In interval notation, the domain of [tex]\( \sec(x) \)[/tex] is:

[tex]\[ (-\infty, -\frac{\pi}{2}) \cup (-\frac{\pi}{2}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \infty) \][/tex]

The formula for volume of cone is:

V = π r^2 h / 3

or

π r^2 h / 3 = 36 cm^3

Simplfying in terms of r:

r^2 = 108 / π h

To find for the smallest amount of paper that can create this cone, we call for the formula for the surface area of cone:

S = π r sqrt (h^2 + r^2)

S = π sqrt(108 / π h) * sqrt(h^2 + 108 / π h) 

S = π sqrt(108 / π h) * sqrt[(π h^3 + 108) / π h] 

Surface area = sqrt (108) * sqrt[(π h + 108 / h^2)] 

Getting the 1st derivative dS / dh then equating to 0 to get the maxima value:

dS/dh = sqrt (108) ((π – 216 / h^3) * [(π h + 108/h^2)^-1/2] 

Let dS/dh = 0 so,

 π – 216 / h^3 = 0 

h^3 = 216 / π

h = 4.10 cm

Calculating for r:

r^2 = 108 / π (4.10)

r = 2.90 cm

Answers:

 h = 4.10 cm

r = 2.90 cm

The height of the cone is [tex]\boxed{4.10}[/tex] and the radius of the cone is [tex]\boxed{2.90}.[/tex]

Further explanation:

The volume of the cone is [tex]\boxed{V = \dfrac{1}{3}\left( {\pi {r^2}h} \right)}.[/tex]

The surface area of the cone is [tex]\boxed{S=\pi \times r\times l}[/tex]

Here l is the slant height of the cone.

The value of the slant height can be obtained as,

[tex]\boxed{l = \sqrt {{h^2} + {r^2}} }[/tex].

Given:

The volume of the cone shaped paper drinking cup is [tex]36{\text{ c}}{{\text{m}}^3}[/tex].

Explanation:

The volume of the cone shaped paper drinking cup  [tex]36{\text{ c}}{{\text{m}}^3}[/tex].

[tex]\begin{aligned}V&=36\\\frac{1}{3}\left({\pi {r^2}h}\right) &= 36\\{r^2}&= \frac{{108}}{{\pi h}}\\\end{aligned}[/tex]

The surface area of the cone is,

[tex]\begin{aligned}S &= \pi\times\sqrt {\frac{{108}}{{\pi h}}}\times\sqrt {{h^2} + \frac{{108}}{{\pi h}}}\\&= \sqrt{108}\times\sqrt{\frac{{\pi {h^3} + 108}}{{\pi h}}}\\&=\sqrt {108}\times\sqrt {\pi h + \frac{{108}}{{{h^2}}}}\\\end{aligned}[/tex]

Differentiate above equation with respect to h.

[tex]\dfrac{{dS}}{{dh}}=\sqrt {108}\times \left( {\pi  - \dfrac{{216}}{{{h^3}}}}\right)\times {\left( {\pi h + \dfrac{{108}}{{{h^2}}}}\right)^{ - \dfrac{1}{2}}}[/tex]

Substitute 0 for [tex]\dfrac{{dS}}{{dh}}[/tex].

[tex]\begin{aligned}\pi- \dfrac{{216}}{{{h^3}}}&= 0\\\dfrac{{216}}{{{h^3}}}&= \pi\\\dfrac{{216}}{{3.14}} &= {h^3}\\h &= 4.10\\\end{aligned}[/tex]

The radius of the cone can be obtained as,

[tex]\begin{aligned}{r^2}&=\frac{{108}}{{\pi \left({4.10} \right)}}\\{r^2}&= \frac{{108}}{{3.14 \times 4.10}}\\{r^2}&= 8.40\\r&= \sqrt {8.40}\\r &= 2.90\\\end{aligned}[/tex]

Hence, the height of the cone is  [tex]\boxed{4.10}[/tex]and the radius of the cone is [tex]\boxed{2.90}[/tex].

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Mensuration

Keywords: cone shaped paper, drinking cup, volume, [tex]36{\text{ c}}{{\text{m}}^3}[/tex], height of cone, cup, smallest amount of paper, water.