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3 years ago

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Ramen noodles. There are 20 ramen noodle packages in a bag. There are 10 beef flavored, and the other 10 are chicken flavored. W

hat is the probability of getting at least one chicken and at least one beef flavor when you grab 3 packages randomly?

1 answer:

Elina [12.6K]3 years ago

8 0

Answer:

0.789

Step-by-step explanation:

Number of beef flavored Ramen noodles = 10

Number of chicken flavored Ramen noodles = 10

Total number of noodles = 20

Number of ways to select 3 packages:

$_{20}C_3= \dfrac{20\times 19\times 18}{3\times 2\times 1}=1140$

Number of ways to select two beef flavored Ramen noodle = $_{10}C_2$ = 45

Number of ways to select one chicken flavored Ramen noodle = 10

Number of ways to select two chicken flavored Ramen noodle = $_{10}C_2$ = 45

Number of ways to select one chicken flavored Ramen noodle = 10

Total number of ways so that one chicken and one beef Ramen selected = 450 + 450 = 900

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

Required probability is:

$\dfrac{900}{1140} = 0.789$

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3 years ago

ASAP 30 + Brainliest <br><br> Please only solve 2 - 5

hichkok12 [17]

<u>QUESTION 2a</u>

We want to find the area of the given right angle triangle.

We use the formula

$Area=\frac{1}{2}\times base\times height$

The height of the triangle is $=a cm$ .

The base is $12cm$ .

We substitute the given values to obtain,

$Area=\frac{1}{2}\times 12\times a cm^2$ .

This simplifies to get an expression for the area to be

$Area=6a cm^2$ .

<u>QUESTION 2b</u>

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=7cm$ and the width of the rectangle is $w=ycm$ .

We substitute the values to obtain the area to be

$Area=7 \times y$

The expression for the area is

$Area=7y$

<u>QUESTION 2c.</u>

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=2x cm$ and the width of the rectangle is $w=4 cm$ .

We substitute the values to obtain the area to be

$Area=2x \times 4$

The expression for the area is

$Area=8x$

<u>QUESTION 2d</u>

The given diagram is a square.

The area of a square is given by,

$Area=l^2$ .

where $l=b m$ is the length of one side.

The expression for the area is

$Area=b^2 m^2$

<u>QUESTION 2e</u>

The given diagram is an isosceles triangle.

The area of this triangle can be found using the formula,

$Area=\frac{1}{2}\times base\times height$ .

The height of the triangle is $4cm$ .

The base of the triangle is $6a cm$ .

The expression for the area is

$Area=\frac{1}{2}\times 6a \times 4cm^2$

$Area=12a cm^2$

<u>QUESTION 3a</u>

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

<u>QUESTION 3b</u>

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

Or

$P=2(L+B)$

<u>QUESTION 3c</u>

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

$Perimeter=3q+P+3q+P$

This simplifies to,

$Perimeter=6q+2P$

Or

$Perimeter=2(3q+P)$

<u>QUESTION 3d</u>

The given figure is a rhombus.

The perimeter is the distance around the whole figure.

Let P be the perimeter. Then

$P=5b+5b+5b+5b$

This simplifies to,

$P=20b mm$

<u>QUESTION 3e</u>

The given figure is an equilateral triangle.

The perimeter is the distance around this triangle.

Let P be the perimeter, then,

$P=2x+2x+2x$

We simplify to get,

$P=6x mm$

QUESTION 3f

The figure is an isosceles triangle so two sides are equal.

We add all the distance around the triangle to find the perimeter.

This implies that,

$Perimeter=3m+5m+5m$

$Perimeter=13m mm$

<u>QUESTION 3g</u>

The given figure is a scalene triangle.

The perimeter is the distance around the given triangle.

Let P be the perimeter. Then

$P=(3x+1)+(2x-1)+(4x+5)$

This simplifies to give us,

$P=3x+2x+4x+5-1+1$

$P=9x+5$

<u>QUESTION 3h</u>

The given figure is a trapezium.

The perimeter is the distance around the whole trapezium.

Let P be the perimeter.

Then,

$P=m+(n-1)+(2m-3)+(n+3)$

We group like terms to get,

$P=m+2m+n+n-3+3-1$

We simplify to get,

$P=3m+2n-1$ mm

QUESTION 3i

The figure is an isosceles triangle.

We add all the distance around the figure to obtain the perimeter.

Let $P$ be the perimeter.

Then $P=(2a-b)+(a+2b)+(a+2b)$

We regroup the terms to get,

$P=2a+a+a-b+2b+2b$

This will simplify to give us the expression for the perimeter to be

$P=4a+3b$ mm.

QUESTION 4a

The given figure is a square.

The area of a square is given by the formula;

$Area=l^2$

where $l=2m$ is the length of one side of the square.

We substitute this value to obtain;

$Area=(2m)^2$

This simplifies to give the expression of the area to be,

$Area=4m^2$

QUESTION 4b

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=5a cm$ is the length of the rectangle and $w=6cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =5a \times 6$

$Area =30a cm^2$

QUESTION 4c

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=7y cm$ is the length of the rectangle and $w=2x cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =7y \times 2x$

The expression for the area is

$Area =14xy cm^2$

QUESTION 4d

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=3p cm$ is the length of the rectangle and $w=p cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =3p \times p$

The expression for the area is

$Area =3p^2 cm^2$

See attachment for the continuation

6 0

3 years ago

Read 2 more answers

BRAINLIESTT ASAP! PLEASE HELP ME :)

Anvisha [2.4K]

Step-by-step explanation:

$\displaystyle y = Asec(Bx - C) + D$

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Therefore we have our answer.

Extended Information on the trigonometric function

$\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|$

NOTE: Sometimes, your <em>vertical shift</em> might tell you to shift your graph below or above the <em>midline</em> where the amplitude is. Moreover, ALL <em>tangent</em>,<em> </em><em>secant</em>, <em>cosecant</em>, and <em>cotangent</em> functions have NO AMPLITUDE.

I am joyous to assist you anytime.

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