Can you guys help me and answer this math question below please I really need help I'll mark brainliest to whoever answers please help

Answer:

b.

Step-by-step explanation:

Complete Question:

If a 100kg bag of rice cost #120, how many kilogram of rice can one buy for #40

Answer:

33 kilograms.

Step-by-step explanation:

Given the following data;

100 kg of rice = #120

1 kg of rice = #x

Cross-multiplying, we have;

100x = 120

x = 120/100

x = 1.2

Now, to find how much #40 can buy;

[tex] Quantity = \frac {Cost}{Cost \; of \; a \; bag} [/tex]

Substituting into the equation, we have;

[tex] Quantity = \frac {40}{1.2} [/tex]

Quantity of rice = 33.33 ≈ 33 kilograms.

Method II

100 kg of rice = #120

x kg of rice = #40

Cross-multiplying, we have;

120x = 40 * 100

x = 4000/120

x = 33.33 ≈ 33 kilograms.

Answer:

83.8

Step-by-step explanation:

first you add the grades together: 75 + 87 + 90 + 88 + 79 = 419

then you divide by the amount of grades: 419 / 5 = 83.8

Answer:

A

Step-by-step explanation:

A is the answer for the first question I did the math on edge

The second one after is B

We are Simplifying Rational Expressions with Negative Exponents

Answer:

A.) Afriv

Step-by-step explanation:

Heat oven to 350°F. Generously grease 12-cup fluted tube pan with shortening or cooking spray. In large 1-gallon plastic food storage bag, mix granulated sugar and cinnamon.

2

Separate dough into 16 biscuits; cut each into quarters. Shake in bag to coat. Arrange in pan, adding walnuts and raisins among the biscuit pieces. Sprinkle any remaining sugar over biscuits.

3

In small bowl, mix brown sugar and butter; pour over biscuit pieces.

4

Bake 30 to 40 minutes or until golden brown and no longer doughy in center. Loosen edges of pan with metal spatula. Cool in pan 5 minutes. Turn upside down onto serving plate; replacing any biscuit pieces and caramel from pan. Pull apart to serve. Serve warm.

Answer:

5

Step-by-step explanation:

The interquartile range is the 3rd quartile minus the 1st quartile. The 3rd quartile is 8 and the 1st quartile is 3.

8 - 3 = 5

The interquartile range (IQR) of the data is 5.

From the box plot we can write,

Minimum = 2

First Quartile, Q1 = 3

Median = 6

Third Quartile, Q3 = 8

Maximum = 10

So, Interquartile range (IQR)

= Q3 - Q1

= 8 - 3

= 5

Thus, the interquartile range (IQR) of the data is 5.

Learn more about Interquartile range (IQR) here:

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Answer:it’s c y=|x-5|-4

Step-by-step explanation:

The correct function which is graphed is,

D. [tex]y = |x- 5| - 4[/tex]

We have to give that,

A graph of a function is shown in the image.

Now, From the graph;

Two points on the function are (0, 1) and (5, 4).

Since both points are lies on the graph of a function, hence it satisfies the equation of the function.

From the given function,

A. [tex]y = |x + 5|+ 4[/tex]

Substitute x = 0, y = 1;

1 = |0 + 5| + 4

1 = 5 + 4

1 = 9

Which is not true.

B. [tex]y = |x -5| + 4[/tex]

Substitute x = 0, y = 1;

1 = |0 - 5| + 4

1 = 5 + 4

1 = 9

Which is not true.

C. [tex]y = |x + 5| - 4[/tex]

Substitute x = 0, y = 1;

1 = |0 + 5| - 4

1 = 5 - 4

1 = 1

Which is true.

And, Substitute x = 5, y = 4;

4 = |5 + 5| + 4

4 = 10 + 4

4 = 14

Which is not true.

D. [tex]y = |x- 5| - 4[/tex]

Substitute x = 0, y = 1;

1 = |0 - 5| - 4

1 = 5 - 4

1 = 1

Which is true.

And, Substitute x = 5, y = 4;

4 = |5 - 5| + 4

4 = 0 + 4

4 = 4

Which is true.

To learn more about the function visit:

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Answer:

c

Step-by-step explanation:

you would do

m = money overall

(m*1.08) +3000= you would do that twelve times

Answer:

no

Step-by-step explanation:

5/8=0.6250 so its rational

Answer: no

Step-by-step explanation: 5/8 is rational because it is another way of writing the ratio 5:8. So no, 5/8 is not an irrational number.

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the derivatives.

Hence, here

y = 6x^2 -4x +17

so, dy/dx = d(6x^2-4x+17)/dx

===> hence, dy/dx = 12x -4 ( used formula d(x^n) /dx = nx^n-1 )

======> dy/dx = 12x-4

Answer:

1.) 7x+5y

2.)-19m+14r

3.)22b+40c

4.)10x+25

5.)7m+7

Step-by-step explanation:

ez

Answer:

The answer is 7/12

Answer:

7/12

Step-by-step explanation:

Answer:

6x=18

x=18/6

x=3.

-3x=15

x=15/-3

x=-5

7x=-63

x=-63/7

x=-9

5x-3=12

5x=12+3

5x=15

x=15/5

x=3

-4x+1=13

-4x=13-1

-4x=12

x=12/-4

x=-3

-x+9=16

-x=16-9

-x=7

x=7/-1

x=-7

Answer:

z = 71

Step-by-step explanation:

this is the answer for this equation

Answer:

z=71

Step-by-step explanation:

Answer:

the surface area of the wedge is 360ft²

Step-by-step explanation:

6x5x3x4 = 360ft²

Answer:

srry buddy I don't understand your language

I am from India

can ask it in English

then I could help you

Answer:

a. area of triangle: A=1/2(b)(h)

                               A= 1/2 (3)(2.4)

                               A= 3.6cm^2

All the triangles have the same area, so Total area of the triangles is

A= 4(3.60)

A= 14.4 cm^2

Area of the square: A= s^2 (3)^2= 9cm^2

Total surface area o the Pyramid:  14.4+ 9=23.4cm^2

b. Lateral surface area of the Pyramid: A= 1/2 P(l)              

                                                                 A= 1/2 (12)(4)

                                                                  A= 24cm^2

 P is perimeter of the base= (3)(4)=12cm

 l s the slant height= 4cm

Step-by-step explanation:

Answer:

The quadratic equation is [tex]h(t) = -318.22t^2 + 954.67t + 12728.89[/tex]

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

[tex]f(x) = ax^{2} + bx + c[/tex]

It's vertex is the point [tex](x_{v}, y_{v})[/tex]

In which

[tex]x_{v} = -\frac{b}{2a}[/tex]

[tex]y_{v} = -\frac{\Delta}{4a}[/tex]

Where

[tex]\Delta = b^2-4ac[/tex]

If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

[tex]ax^{2} + bx + c, a\neq0[/tex].

This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:

[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]

[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]

[tex]\Delta= b^{2} - 4ac[/tex]

Quadratic equation:

[tex]ax^2 + bx + c = 0[/tex]

The rock reaches a maximum height of 676 feet above the ground after 1.5 seconds.

This means that:

[tex]-\frac{b}{2a} = 1.5[/tex]

[tex]y_{v} = -\frac{b^2-4ac}{4a} = 676[/tex]

If the rock hits the ground below the ledge 8 seconds after it is thrown

This means that:

[tex]64a + 8b + c = 0[/tex]

Relation of b and a:

[tex]-\frac{b}{2a} = 1.5[/tex]

This means that:

[tex]b = -3a[/tex]

Relationship of c and a:

[tex]64a + 8b + c = 0[/tex]

[tex]64a - 24a + c = 0[/tex]

[tex]c = -40a[/tex]

Finding a:

[tex]y_{v} = -\frac{(-3a)^2-4a(-40a)}{4a} = 676[/tex]

[tex]9a^2+160a = -2704a[/tex]

[tex]9a^2 + 2864a = 0[/tex]

[tex]a(9a + 2864) = 0[/tex]

Since it is a quadratic equation, a cannot be 0.

[tex]9a + 2864 = 0[/tex]

[tex]9a = -2864[/tex]

[tex]a = -\frac{2864}{9}[/tex]

[tex]a = -318.22[/tex]

Finding b and c:

[tex]b = -3a = -3(-318.22) = 954.67[/tex]

[tex]c = -40a = -40(-318.22) = 12728.89[/tex]

The quadratic equation is given by:

[tex]h(t) = -318.22t^2 + 954.67t + 12728.89[/tex]

The quadratic equation is [tex]h(t) = -318.22t^2 + 954.67t + 12728.89[/tex]

Answer:

11

Step-by-step explanation:

First one: 7

Second one: 22

Third one: 137

Hope this helps.

Answer:

Step-by-step explanation:

Two of the sides are 13.5 inches by 2.5 inches, two sides are 13.5 inches by 8 inches, and two sides are 2.5 inches by 8 inches.

Surface area = 2*13.5*2.5 + 2*13.5*8 + 2*2.5*8

Step-by-step explanation:

so basically they are asking you to subtract, hope this helps, use a calculator if needed

Answer:

-3 ≥ b is your inequality

Step-by-step explanation:

Answer:

123

Step-by-step explanation:

213

Answer:

5x-5y

Step-by-step explanation:

You are using the distributive property

You can also think of it as multiplying 5 by x and 5 by y and then subtracting it just as it is subtracting in the original equation

Answer:

5x-y.

Step-by-step explanation:

so its 5 times x, minus 7. easy. 5x,-y. rephrase it. hhen a variable is next to a number, it means multiply.

Answer:

The value for x should be 35

Step-by-step explanation:

Answer:

32 degrees

Step-by-step explanation:

35 + x + 29 + 84 = 180

148 + x = 180

x = 32

Answer: 400 because when they test 200 they get that 4 are dead wich is 1/50 of 200 so 1/50 times 20,000 is 400

Step-by-step explanation:

Answer:

3.6c + 7.5t <= 21,600

Step-by-step explanation:

Let c = number of cars, and let t = number of trucks.

c cars weigh 3.6c tons, and t trucks weigh 7.5t tons.

3.6c + 7.5t <= 21,600

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

[tex]\to f_X (x) \ \ xe^{-x} \ , \ x>0[/tex]

For point a:

Moment generating function of X=?

Using formula:

[tex]\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx[/tex]

[tex]M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx[/tex]

integrating the values by parts:

[tex]u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\[/tex]  

        [tex]= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\[/tex]

Therefore, the moment value generating by the function is [tex]=\frac{1}{(t-1)^2}[/tex]

In point b:

[tex]E(X^n)=?[/tex]

Using formula: [tex]E(X^n)= M^{n}_{X}(0)[/tex]

form point (a):

[tex]\to M_{X}(t)=\frac{1}{(t-1)^2}[/tex]

Differentiating the value with respect of t

[tex]M'_{X}(t)=\frac{-2}{(t-1)^3}[/tex]

when [tex]t=0[/tex]

[tex]M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\[/tex]

[tex]M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\[/tex]

Step-by-step explanation:

(8+(-2)-4)

=8-2-4

=2

(3-5)

=-2