A cup holds 2\9 litre of water. How many cups of water can be filled with 3litre bottle

The laterals surface area of the given hexagonal prism is: 250 cm².

Lateral surface area = 5ah, where:

Given the following:

a = 5 cm

h = 10 cm

Lateral surface area of the hexagonal prism = 5ah =  5(5)(10)

Lateral surface area of the hexagonal prism = 250 cm²

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Answer: 9

Step-by-step explanation:

Answer:

9 songs

Step-by-step explanation:

a quarter is basically a fourth of a number so you need to just divide the number by four

The values that are included in the random variable of the experiment that include rolling one 6-sided die and one 10-sided die are 10, 14 and 16

The experiment elements are given as:

The sample spaces of both dice are:

6-sided = {1, 2, 3...., 6}

10-sided = {1, 2, 3...., 10}

When the outcomes of both dice are added, we have:

Sample space = {2, 3, 4, 5, 6, 7, 8, 9, .............. 16}

Minimum = 2

Maximum = 16

This means that the values that are included in the random variable are in the range 2 - 16 i.e. starting from 2 and ending in 16

Hence, the values that are included in the random variable of using one 6-sided die and one 10-sided die and rolling them together are 10, 14 and 16

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The area of the actual building lot is 5040 square ft

This factors are used to enlarge or diminish the given figure. Given the  scale model of a rectangular building lot measures 7 feet by 5 feet. If the actual house will be built using a scale factor of 12, the new length and width is expressed as:

Length = 12 * 7 = 84ft

Width = 5*12=60 ft

Determine the area

Area = 84 * 60

Area = 5040 square feet

Hence the area of the actual building lot is 5040 square ft

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The measure of angle BC (m ∠BC) is 90°

From the question, we are to determine the measure of angle BC

From the given information, we have that

B P ⊥ P C

This means line BP is perpendicular to line PC

The angle between two perpendicular lines is 90°

Hence, the measure of angle BC (m ∠BC) is 90°.

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1003 more square feet of wallpaper for the living room than the family room when family room is with dimensions (l = 3x, b = 3x, h = x), living room is with dimensions (l = 5x, b = 3x, h = x) and doors are with dimension (b =3,    h = 7) {where x = 8}. This can be obtained by forming equations for the given figures.

Given that,

dimensions (l = 3x, b = 3x, h = x) and there is one door (b =3, h = 7)

expression of square feet of wallpaper for family room,

⇒ area of four walls - area of one door= [(3x)(x)+(3x)(x)+(3x)(x)+(3x)(x)] - [3×7]

                                                               =[3x²+3x²+3x²+3x²] - 21

                                                               =12x² - 21 square feet

Given that,

height and width are same as of the family room but length is 5 times more than height [length = 5(height) ⇒ l = 5x]

dimensions (l = 5x, b = 3x, h = x) and there is two doors (b =3, h = 7)

expression of square feet of wallpaper for family room,

area of four walls - area of two doors= [(3x)(x)+(5x)(x)+(3x)(x)+(5x)(x)] - 2[3×7]

                                                               =[3x²+5x²+3x²+5x²] - 2(21)

                                                               =16x² - 42 square feet

total amount of wallpaper needed

⇒wallpaper for family room + wallpaper for living room = 12x²-21 + 16x²-42

                                                                                        =28x²-63 square feet

Area of wallpaper of living room - area of wallpaper of family room

= 1750 - 747

= 1003 square feet

Hence 1003 more square feet of wallpaper for the living room than the family room when family room is with dimensions (l = 3x, b = 3x, h = x), living room is with dimensions (l = 5x, b = 3x, h = x) and doors are with dimension (b =3,    h = 7) {where x = 8}.

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The number of ways we can select 3 cracked and 2 good is 1200 ways.

Given there are 5 eggs in which 3 eggs are cracked and 2 eggs which are good.

We have to find the number of ways we can select 3 cracked and 2 good eggs from 5 eggs.

Number of ways can be found through permutations.

The factorial of n  equals the product of n with the smaller factorial..

Permutations is the technique of calculating  various ways in which objects from a set may be selected, generally without replacement, to form subsets.

n[tex]P_{r}[/tex]=n!/(n-r)!

So the number of ways will be 5[tex]P_{3}[/tex]*5[tex]P_{2}[/tex]

([tex]5P_{3}[/tex] for cracked eggs and [tex]5C_{2}[/tex] for good eggs)

=5!/(5-3)!*51/(5-2)!

=5*4*3*2!/2!*5*4*3!/3!

=5*4*3*5*4

=1200

Hence there are 1200 ways through which 3 cracked and 2 good eggs can be selected from 5 eggs.

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The expression that represents well the geometric sequence is [tex]f(n) = 8 \cdot 2^{n-1}[/tex] and the value of the fourth term is 64. (Correct choice: B)

Geometric sequences are exponential expressions with discrete domain, whose form is presented and explained below:

[tex]f(n) = a \cdot r^{n-1}[/tex], where a, r are real numbers and n is a natural number.     (1)

Where:

According to the statement, we know that the first term of the geometric sequence is 8 and between any two consecutive terms there is a common ratio of 2. If we know that a = 8, r = 2 and n = 4, then the fourth term of the series by means of (1) is:

[tex]f(4) = 8 \cdot 2^{4-1}[/tex]

f(4) = 8 · 8

f(4) = 64

The expression that represents the geometric sequence is [tex]f(n) = 8 \cdot 2^{n-1}[/tex] and the value of the fourth term is 64.

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The measure of angle E is (b) 50°

The given parameters are:

The similarity of the triangles implies that:

From the figure, we have:

angle A = 50

Substitute angle A = 50 in angle A = angle E

angle E = 50

Hence, the measure of angle E is (b) 50°

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Using the normal distribution, we have that:

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The mean and the standard deviation for this problem are given, respectively, by:

[tex]\mu = 5.34, \sigma = 0.03[/tex]

The 8th percentile separates the bottom 8%, that is, X when Z = -1.405, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.405 = \frac{X - 5.34}{0.03}[/tex]

X - 5.34 = -1.405 x 0.03

X = 5.30.

The 92th percentile separates the top 8%, that is, X when Z = 1.405, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.405 = \frac{X - 5.34}{0.03}[/tex]

X - 5.34 = 1.405 x 0.03

X = 5.38.

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What is one major limitation of the Social Readjustment Rating Scale?

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Answer: B

[tex]x + 2y \leq 4 \textrm{ and } 3x + y \leq 4[/tex]

Step-by-step explanation:

Insert the values of x and y for any point in the heavily shaded region and see if it satisfies both inequalities

We can see that the origin (0,0) is inside the shaded region so use these values because they are the easiest to compute

Plug in x = 0 and y = 0 into the first inequality
0 + 2.0 = 0 which is greater than 5.

So we can eliminate both A and D since they both contain incorrect inequalities

Now, test the second inequality with x = 0 and y = 0

For B we get

3.0 + 0 = 0 which is indeed less than 4

C is not applicable since 0 is not greater-than-equal to 4

Hence B is the correct solution

This can be verified using a graphing tool (see image)

The area of the patio is 192 [tex]m^{2}[/tex]

An object's area is how much space it takes up in two dimensions. It is the measurement of the quantity of unit squares that completely cover the surface of a closed figure. The square unit, which is frequently expressed as square inches, square feet, etc., is the accepted unit of area.

The word "area" refers to a free space. A shape's length and width are used to compute its area.

Length of swimming pool=18 m

Width of swimming pool=8 m

Area of swimming pool= 18*8 = 144[tex]m^{2}[/tex]

Width of patio=3m

Total length of swimming pool=18+2(3)=24 m

Total width of swimming pool=8+2(3)=14 m

Total area=24*14=336 [tex]m^{2}[/tex]

Area of patio= total area - Area of swimming pool

                     =336-144

                     =192 [tex]m^{2}[/tex]

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The  appropriate display to show the spread of 150 test scores for Mrs. Hansen's last science test is: b.  box and whisker plot.

Box and whisker plot which is also called box plot can be defined as the plot that show a graphical presentation of data set by helping to display the data set or data score .

Hence a graphical presentation display to show the spread of 150 test scores for Mrs. Hansen's last science test is option B which is Box and whisker plot .

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The concept of factoring quadratic equations simply is one of the ways to solve for the zeroes of the quadratic equation.

The concept of factoring quadratics is a technique which is characterized by expressing the quadratic equation ax2 + bx + c = 0 as a product of its associated linear factors as (x - f)(x - g), in which case, f and g are the roots of the quadratic equation ax2 + bx + c = 0.

Factoring is an important concept that helps express equations of any type in simpler forms The mode of operation of the factoring works such that, the polynomial is rewritten in a simpler form and consequently, expressed as a product of it's linear factors.

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The length of the shortest side is DC

As a general rule in triangle;

The smallest angle in a triangle is opposite the shortest side.

Similarly, the largest angle in a triangle is opposite the largest side.

From the given figure of triangles, the smallest angle is angle CBD with a measure of 47 degrees.

The side opposite this angle is side length DC

Hence, the length of the shortest side is DC

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The wonderful waffle cone holds more ice cream than super sundae cone.

In this question,

The ice cream cone is the shape of hemisphere at the top and cone shape at the bottom.

Volume of ice cream cone = volume of hemisphere + volume of cone

Volume of ice cream cone = [tex]\frac{2}{3}\pi r^{3} +\frac{1}{3} \pi r^{2} h[/tex]

Volume of super sundae cone:

Radius of hemisphere = height of hemisphere = 0.5 in

Depth of the ice cream cone = 4 in

Volume of super sundae cone = [tex]\frac{2}{3}\pi (0.5)^{3} +\frac{1}{3} \pi (0.5)^{2} (4)[/tex]

⇒ [tex]\frac{2}{3}(3.14)(0.125) +\frac{1}{3} (3.14)(0.25)(4)[/tex]

⇒ 0.2616+1.0466

⇒ 1.3082 ≈ 1.31 cubic in.

Volume of wonderful waffle cone:

Radius of hemisphere = height of hemisphere = 0.9 in

Depth of the ice cream cone = 3 in

Volume of wonderful waffle cone = [tex]\frac{2}{3}\pi (0.9)^{3} +\frac{1}{3} \pi (0.9)^{2} (3)[/tex]

⇒ [tex]\frac{2}{3}(3.14)(0.729) +\frac{1}{3} (3.14)(0.81)(3)[/tex]

⇒ 1.526+2.5434

⇒ 4.0694 ≈ 4.07 cubic in.

Thus volume of wonderful waffle cone > volume of super sundae cone.

Hence we can conclude that the wonderful waffle cone holds more ice cream than super sundae cone.

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Answer:

{x:x is a real number}

Since any value can replace x to give an integer.

{x:x is a real number}

It's kind of self-evident. Any power of 21 will have a 1 in the units place, and adding 9 to it makes it 0 and hence the number is divisible by 5.

To prove the claim by induction, first establish the base case. I assume [tex]n\in\Bbb N[/tex], so for [tex]n=1[/tex] we have

[tex]21^1 + 9 = 21 + 9 = 30[/tex]

and of course 30 = 5×6 is divisible by 5.

Assume the claim is true for [tex]n=k[/tex], that [tex]5 \mid 21^k + 9[/tex]. This means that for some integer [tex]\ell[/tex], we can write

[tex]21^k + 9 = 5\ell[/tex]

Now the induction step: when [tex]n=k+1[/tex], we have

[tex]21^{k+1} + 9 = \bigg(21\times(21^k + 9) - 9\times21\bigg) + 9 \\\\ ~~~~~~~~~~~~~~ = 21\times5\ell - 9\times20 \\\\ ~~~~~~~~~~~~~~= 5\times(21\ell - 9\times4)[/tex]

which is divisible by 5. QED

The interquartile range of the data set is 8

To do this we have to arrange in ascending order

= 8,9,9,9,10,11,13,15,17,18,22

Q2 = The median of the data is 11.

We have to find the median of the half before 11 and the median of the half after 11.

We would have

8,9,9,9,10

Q1 = 9

And 13,15,17,18,22

Q3 = 17

The interquartile range is

Q3 - Q1

= 17 - 9

= 8

The interquartile range of the data set is 8

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Correct question

Study the following data set.

{8,15,9,18,9,17,22,10,11,9,13}

What is the interquartile range of the data set?

The perimeter of the triangle whose coordinates are A (2, 2), B (3, 3), and C (4, 2) is 4.828

We can determine the perimeter of a triangle with vertices by utilizing the distance formula.

Given that, we have the vertices (x₁,y₁), (x₂, y₂), (x₃, y₃). Then the length of each sides can be expressed as follows:

SIde AB = [tex]\mathbf{\sqrt{(x_1 -x_2)^2 +(y_1 -y_2)^2}}[/tex]

Side AC = [tex]\mathbf{\sqrt{(x_1 -x_3)^2 +(y_1 -y_3)^2}}[/tex]

Side BC = [tex]\mathbf{\sqrt{(x_2 -x_3)^2 +(y_2 -y_3)^2}}[/tex]

Replacing the values of the coordinates into the given formula above the perimeter of the triangle is 4.828

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The slope of given data is a -3.

According to the statement

we have given a some data and we have to find the slope of the data.

So, data given is:

x, has the entries, negative 2, negative 1, 0, 1, 2.

y, has the entries, negative 2, 1, 4, 7, 10.

and we have to find the slope of the data

So, we know that it is a data represent in the graph then

We know that a there is a graphical representation of data.

A graph is the representation of the data on the vertical and horizontal coordinates so we can see the trend of the data.

And when we observe this data then we see that the there is some increase the y axis value as compare to the x axis values.

So,

For each unit of run in the x-values, there is an increase of 3 in the y-values. The slope is rising over Run so 3/1 = 3

So, The slope of given data is a -3.

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Answer: the answer is a -3

Step-by-step explanation:

False, Whales evolved in the early Cambrian and include transitional species such as Kutchicetus and Rodhocetus.

What animals evolved during Cambrian period?

When did animals evolve in the Cambrian?

What long lived group of arthropods went extinct at the end of the Permian?

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The complete question is -

True or false . Whales evolved in the early Cambrian and include transitional species such as Kutchicetus and Rodhocetus.

Answer:

Step-by-step explanation:

The amplitude of a sine function is equal to one-half the distance between the maximum and minimum values of the function.

In this case, the maximum value is approximately 84 customers, and the minimum value is approximately 30 customers

Therefore, the value that is closest to the amplitude of the transformed function is 27 customers.

The sum of the numbers is 7.69 * 10^6

The expression is given as:

(1.54 times 10 superscript 6 baseline) (6.15 times 10 superscript 6 baseline)

Rewrite properly

(1.54 * 10^6) + (6.15 * 10^6)

Factor out 10^6

(1.54 + 6.15) * 10^6

Evaluate the sum

7.69 * 10^6

Hence, the sum of the numbers is 7.69 * 10^6

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Answer:

a

Step-by-step explanation:

Find the sum. Express the answer in scientific notation.

(1.54 times 10 Superscript 6 Baseline) + (6.15 times 10 Superscript 6 Baseline)

7.69 times 10 Superscript 6

9.471 times 10 Superscript 6

7.69 times 10 Superscript 12

9.471 times 10 Superscript 12

Answer:

y = 2/5x -4

Step-by-step explanation:

Point-Slope Form formula

y - y1 = m (x - x1)

slope = m = 2/5

(-5, -6)

y - -6 = 2/5 (x - -5)

y + 6 = 2/5x + 2

y = 2/5x - 4

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The weight of a 36-inch piece of PVC pipe is 0.20 pounds

The given parameters about the 36-inch piece of PVC pipe are

Height, h = 36 inches

Outside diameter, D = 0.82 inches

Inside diameter, d= 0.75 inches

The volume of the PVC pipe is then calculated as:

V = πh(D/2)^2 - πh(d/2)^2

Substitute the known parameters in the above equation

V = 3.14 * 36 * (0.82/2)^2 - 3.14 * 36 * (0.75/2)^2

Evaluate the quotients

V = 3.14 * 36 * 0.41^2 - 3.14 * 36 * 0.375^2

Evaluate the exponents

V = 3.14 * 36 * 0.1681 - 3.14 * 36 * 0.140625

Evaluate the products

V = 19.002024 - 15.89625

Evaluate the difference

V = 3.105774 cubic inches

From the question, we have:

Weight =  0.063 pounds per cubic inch

So, the total weight is

Total weight = 3.105774 cubic inches * 0.063 pounds per cubic inch

Evaluate the product

Total weight = 0.195663762 pounds

Approximate

Total weight = 0.20 pounds

Hence, the weight of a 36-inch piece of PVC pipe is 0.20 pounds

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The total surface area of a regular square pyramid having slant height of 10units and base length of 6 units is 168 square inches which is fifth option.

Given that slant height is 10 units and base lengths of 6 units.

We are required to calculate total surface area of regular square pyramid.

We know that total surface area of regular square based pyramid is equal to [tex]s^{2} +s\sqrt{s^{2} +4h^{2} }[/tex] in which s is side of base length and h is slant height of regular square pyramid.

Total surface area=[tex]6^{2} +6\sqrt{6^{2} +4*10^{2} }[/tex]

=36+6[tex]\sqrt{36+400}[/tex]

=36+6[tex]\sqrt{436}[/tex]

=36+6*20.88

=36+125.28

=161.28

Nearest option is 168 square inches.

Hence the total surface are of regular square based pyramid is option 5 which is 168 square inches.

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