Amelia is calculating the density of soccer balls in a bag. She knows the number of balls in the bag and the volume of the bag. Which of the following formulas can be used to calculate the density of balls in the bag?

Density = volume of bag over number of balls
Volume of shelf = density over number of balls
Density = number of balls over volume of bag
Number of books = density over volume of bag

Answer:

hello,

a) answer: 600

b) answer: 840

Step-by-step explanation:

a)

20=2²*5,25=5²,30=2*3*5,40=2³*5

l.c.m(20,25,30,40)=2³*3*5²=8*3*25=800

b)

24=2³*3, 42=2*3*7,35=5*7

l.c.m=(24,42,35)=2³*3*5*7=840

(256⁴-1)

= (256-1)⁴

Using identity (a-b)⁴ = a⁴−4a³b+6a²b²−4ab³+b⁴

Let a be 256 and b be 1

Then

256⁴−4(256)³(1)+6(256)²(1)²−4(256)(1)³+(1)⁴

After solving

(256²-1)²

(a-b)² = a²-2ab+b²

256²-2×256×1+1²

= (256²-1)(256²+1)

Must click thanks and mark brainliest

Answer:

Use identity:

Consider that:

Now factorize:

Answer:

3-4/5=2.2

hope it helps

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]5\frac{7}{10}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Calculating the answer...}}\\\\6\frac{1}{2}-\frac{4}{5} \\-------------\\6\frac{1}{2} = \frac{13}{2} \\\\\frac{13}{2} - \frac{4}{5}\\\\LCM(2,5): 10\\\\\frac{13}{2} =\frac{13*5}{2*5} =\frac{65}{10}\\\\\frac{4}{5}=\frac{4*2}{5*2}=\frac{8}{10}\\\\\frac{65}{10}-\frac{8}{10}\\\\ \frac{57}{10}\\\\\frac{57}{10}\rightarrow\boxed{5\frac{7}{10}}[/tex]

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Step-by-step explanation:

here's the answer to your question

Answer:

the answer to your question is 7.8 (E)

The  distance from point N to LM is 7.8, 8.1 and 8.4 unit.

Perpendicular lines are those that cross at a straight angle to one another. Examples include the opposite sides of a rectangle and the steps of a straight staircase. the icon used to represent two parallel lines.

Perpendicular lines are two separate lines that cross one other at a right angle, or a 90° angle.

Given:

In ΔNOM

The Perpendicular distance is 7.8 unit

and, Hypotenuse distance id 8.1 unit

Now, In ΔNOL

The Perpendicular distance is 7.8 unit

and, Hypotenuse distance id 8.4 unit

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The expected value of the distribution of the number of selected red balls is 0.795.

The expected value of the distribution is the mean or average of the possible outcomes.

There are 12 balls in an urn, five of which are crimson. The selection of a red ball is desired and hence considered a success.

In this case, the possible outcomes are 0, 1, 2, or 3 red balls.

To calculate the expected value, we need to find the probability of each outcome and multiply it by the value of the outcome.

The probability of selecting 0 red balls is :

[tex]$\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{105}{660}$[/tex].

The probability of selecting 1 red ball is :

[tex]$3 \cdot \frac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} + 3 \cdot \frac{7}{12} \cdot \frac{5}{11} \cdot \frac{6}{10} + 3 \cdot \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{315}{660}$[/tex].

The probability of selecting 2 red balls is

[tex]:$\dfrac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} + \frac{7}{12} \cdot \frac{5}{11} \cdot \frac{6}{10} + \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{105}{660}$.[/tex]

The probability of selecting 3 red balls is

[tex]$\dfrac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{15}{660}$[/tex]

The expected value is then :

[tex]$0 \cdot \frac{105}{660} + 1 \cdot \frac{315}{660} + 2 \cdot \frac{105}{660} + 3 \cdot \frac{15}{660} = \frac{525}{660} = \frac{175}{220} \approx \boxed{0.795}$[/tex]

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Answer:

TU = 27

Step-by-step explanation:

We are given two secant segments that are drawn from a circle to meet at an exterior point of the circle. Thus, according to the secant secant theorem, the product of the measure of one secant segment and its external secant segment equals that of the product of the other and its external secant segment.

Thus:

VU*TU = VW*BW

Substitute

7(x + 4) = 9(-2 + x)

7x + 28 = -18 + 9x

Collect like terms

7x - 9x = -18 - 28

-2x = -46

Divide both sides by -2

x = -46/-2

x = 23

✔️TU = x + 4

Plug in the value of x

TU = 23 + 4

TU = 27

Answer:

-1.25

Step-by-step explanation:

you first have to cross multiply

12(b-1)=4(7b+2)

12b-12=28b+8

group the like terms

12b-28b=8+12

-16b/-16=20/-16

b= -1.25

I hope it helps

Step-by-step explanation:

Answer is in the picture..

hope it helps

Answer:

Part A: y = 25x + 30

Part B: $180

Step-by-step explanation:

for part A, 25 is the constant so that goes with the x, and you just add 30 because it is the extra fee. slope intercept form is y = mx + b

for part B you just plug 6 into the slope intercept equation

y = 25(6) + 30

y = 150 + 30

y = 180

$180

Answer:

I think i don't know the answer i am so sorry!!!

maybe someone else can Answer

9514 1404 393

Answer:

Step-by-step explanation:

The function is defined for all values of x, so its domain is all real numbers.

The function can produce values of f(x) that are 0 or greater, so its range is ...

  y ≥ 0

Answer:

0

Step-by-step explanation:

Arc sec(1)=0, tan(0)=0

Answer:

The slope is -4.

Step-by-step explanation:

Slope(m)=(y2-y1)/(x2-x1)

y2=-7, y1=-19, x2=-2, x1=1

(-7+19)/(-2-1)

=12/-3

=-4

Answer: -4

Step-by-step explanation:

The slope formula is: [tex]y_{2} -y_{1}/x_{2}-x_{1} \\[/tex]

So it is: (-7+19)/(-2-1) = 12/-3 = -4

I hope this helped!

Answer:

C.

Step-by-step explanation:

A P E X

please where is the question

Since this is a right triangle, we can use one of the three main trigonometric functions: sine, cosine, or tangent.

Remember: SOH-CAH-TOA

Looking from the given angle, we know the opposite side and want to know the hypotenuse. Therefore, we should use the sine function.

sin(54) = 16/AB

AB = 16/sin(54)

AB = 19.78 units

Hope this helps!

[tex]\begin{array}{c|c|c|c|c|c} p & q & r & p\land q & q\land \neg p & r \land q \\&&&&\\ T & T & T & T & F & T \\ T & T & F & T & F & F \\ T & F & T & F & F & F \\ T & F & F & F & F & F \\ F & T & T & F & T & T \\ F & T & F & F & T & F \\ F & F & T & F & F & F \\ F & F & F & F & F & F\end{array}[/tex]

An implication A => B is true if either A is false, or both A and B are true. So

[tex]\begin{array}{c|c|c}p\land q & (q\land\neg p) \implies (r\land q) & (p\land q) \implies \big[(q\land\neg p) \implies (r\land q)\big] \\&&\\T & T & \mathbf T\\T & T & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\\F & F & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\end{array}[/tex]

and the given statement is a tautology.

9514 1404 393

Answer:

  C.  8

Step-by-step explanation:

The median from vertex B is the line segment between there and the midpoint of side AC. That midpoint is ...

  D = (A +C)/2 = ((-6, 7) +(-2, -9))/2 = (-8, -2)/2 = (-4, -1)

So, we want the length of the line between (-4, -1) and (4, -1). These points are on the same horizontal line (y=-1), so the length is the difference of the x=coordinates:

  median AD = 4 -(-4) = 8 . . . . units in length

Answer:

(x - 4)² + (y + 9)² = 25

Step-by-step explanation:

The equation of a circle is written as seen below.

(x – h)² + (y – k)² = r²

Where (h,k) represents the center of the circle and r represents the radius

We want to find the equation of a circle that has a center at (4,-9) and a radius of 5.

We know that (h,k) represents the center so h = 4 and k = -9

We also know that r represents the radius so r = 5

Now to find the equation of this specific circle we simply plug in these values into the equation of a circle formula

Equation: (x – h)² + (y – k)² = r²

h = 4, k = -9 and r = 5

Plug in values

(x - 4)² + (y - (-9))² = 5²

5² = 25

The two negative signs in front of the 9 cancel out and it changes to + 9

The equation of a circle with a center at (4,-9) and a radius of 5 is

(x - 4)² + (y + 9)² = 25

Answer:

26 + 3 x 42 + 7 x -2 = 138

Step-by-step explanation:

Ok bud, first step we must convert our symbols (Makes it easier to solve)

26 + 3 x 42 + 7 x -2

* subsitutes for multiplication.

I recommend using PEMDAS at times:

1 - Parentheses

2 - Exponents and Roots

3 - Multiplication

4 - Division

5 - Addition

6 - Subtraction

Yet again your numbers were spaced out could they be exponents? if so:

3x^{42}+7x+24

Our answer would round to 24 but he equation was not put in a valid or straight forward way.

Answer:

The probability of getting two numbers whose sum is 10 is 25%.

Step-by-step explanation:

Given that a single die is rolled twice, and there are 36 equally-likely outcomes, to find the probability of getting two numbers whose sum is 10 the following calculation must be performed:

1 = +9

2 = +8

3 = +7

4 = +6

5 = +5

6 = +4

7 = +3

8 = +2

9 = +1

9/36 = 0.25

Therefore, the probability of getting two numbers whose sum is 10 is 25%.

Answer:

142

Step-by-step explanation:

If the enrollment is five times as big, then that is six different things. So, take 852 and divide it by 6 to get 142. Or in other words:

852÷6=142

142x6=856

I hope this helps. Cheers^^

Answer:

30 Dollars more

Step-by-step explanation:

This summer he earned = 7.25 X 40 = 290

Last summer he earned = 6.5 X 40 = 260

How much more he earned in 40 hours = 290-260= 30 dollars more

Answer from Gauthmath

Answer:

(a) 4 sample points

(b) See attachment for tree diagram

(c) The probability that no tail is appeared is 1/4

(d) The probability that exactly 1 tail is appeared is 1/2

(e) The probability that 2 tails are appeared is 1/4

(f) The probability that at least 1 tail appeared is 3/4

Step-by-step explanation:

Given

[tex]Coins = 2[/tex]

Solving (a): Counting principle to determine the number of sample points

We have:

[tex]Coin\ 1 = \{H,T\}[/tex]

[tex]Coin\ 2 = \{H,T\}[/tex]

To determine the sample space using counting principle, we simply pick one outcome in each coin. So, the sample space (S) is:

[tex]S = \{HH,HT,TH,TT\}[/tex]

The number of sample points is:

[tex]n(S) = 4[/tex]

Solving (b): The tree diagram

See attachment for tree diagram

From the tree diagram, the sample space is:

[tex]S = \{HH,HT,TH,TT\}[/tex]

Solving (c): Probability that no tail is appeared

This implies that:

[tex]P(T = 0)[/tex]

From the sample points, we have:

[tex]n(T = 0) = 1[/tex] --- i.e. 1 occurrence where no tail is appeared

So, the probability is:

[tex]P(T = 0) = \frac{n(T = 0)}{n(S)}[/tex]

This gives:

[tex]P(T = 0) = \frac{1}{4}[/tex]

Solving (d): Probability that exactly 1 tail is appeared

This implies that:

[tex]P(T = 1)[/tex]

From the sample points, we have:

[tex]n(T = 1) = 2[/tex] --- i.e. 2 occurrences where exactly 1 tail appeared

So, the probability is:

[tex]P(T = 1) = \frac{n(T = 1)}{n(S)}[/tex]

This gives:

[tex]P(T = 1) = \frac{2}{4}[/tex]

[tex]P(T = 1) = \frac{1}{2}[/tex]

Solving (e): Probability that 2 tails appeared

This implies that:

[tex]P(T = 2)[/tex]

From the sample points, we have:

[tex]n(T = 2) = 1[/tex] --- i.e. 1 occurrences where 2 tails appeared

So, the probability is:

[tex]P(T = 2) = \frac{n(T = 2)}{n(S)}[/tex]

This gives:

[tex]P(T = 2) = \frac{1}{4}[/tex]

Solving (f): Probability that at least 1 tail appeared

This implies that:

[tex]P(T \ge 1)[/tex]

In (c), we have:

[tex]P(T = 0) = \frac{1}{4}[/tex]

Using the complement rule, we have:

[tex]P(T \ge 1) + P(T = 0) = 1[/tex]

Rewrite as:

[tex]P(T \ge 1) = 1-P(T = 0)[/tex]

Substitute known value

[tex]P(T \ge 1) = 1-\frac{1}{4}[/tex]

Take LCM

[tex]P(T \ge 1) = \frac{4-1}{4}[/tex]

[tex]P(T \ge 1) = \frac{3}{4}[/tex]

Answer:

The mean of these numbers is 15.68816 with a repeating 6.

Step-by-step explanation:

Answer:

Centre is,

((5+7)/2,(8+6)/2)

or, (12/2,14/2)

or, (6,7)

radius is,

[√{(5-7)²+(8-6)²}]/2

= [√(2²+2²)]/2

= [√(4+4)]/2

= [√8 ]= [2√2]/2 = √2

The center of a circle = (6, 7)

the radius of a circle = [tex]\sqrt{2}[/tex] units

"The distance between two points [tex](x_1,y_1),(x_2,y_2)[/tex] is given by  [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_2)^2}[/tex]"

"The coordinates of the midpoint of the line segment having endpoints [tex](x_1,y_1),(x_2,y_2)[/tex] is given by [tex]m=(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )[/tex] "

"It is the line segment through the center and touching two points on its edge. "

For given question,

The diameter of circle has endpoints at (5, 8) and (7,6).

Let [tex](x_1,y_1)=(5,8),(x_2,y_2)=(7,6)[/tex]

First we find the length of the diameter of the circle.

Using the distance formula,

[tex]\Rightarrow d=\sqrt{(x_2-x_1)^2+(y_2-y_2)^2} \\\\\Rightarrow d=\sqrt{(7-5)^2+(6-8)^2} \\\\\Rightarrow d=\sqrt{2^2+(-2)^2}\\\\\Rightarrow d=\sqrt{4+4}\\\\\Rightarrow d=2\sqrt{2}~units[/tex]

We know that the diameter = 2 × radius

⇒ radius (r) = [tex]\frac{2\sqrt{2} }{2}[/tex]

⇒ r = [tex]\sqrt{2}[/tex] units

We know that the midpoint of the diameter is the center of the circle.

Using midpoint formula the center of the circle would be,

[tex]\Rightarrow O=(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )\\\\\Rightarrow O=(\frac{5+7}{2} ,\frac{8+6}{2} )\\\\\Rightarrow O=(\frac{12}{2} ,\frac{14}{2} )\\\\\Rightarrow O=(6,7)[/tex]

Therefore, the center of the circle is (6,7)

Hence, the center of a circle = (6, 7)

the radius of a circle = [tex]\sqrt{2}[/tex]

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Answer:

(0,4)  will be point (P) at 3 because,

Step-by-step explanation:

by using newton interpolation method we can find P(0,4) at 3 .

Solution :

Defining the following events as :

B : Being a Business major

α : Studying at the library

∴ Given that :

[tex]$P(B) = \frac{45}{100}$[/tex]

         = 0.45

Again, P [ Studying at the library | Being a Business major ] = 2 P [ Studying at the library | Being a non business major ]

[tex]$P[ \alpha | B] = 2 P[\alpha | B^C]$[/tex]   .......(1)

Again,

[tex]$P[\text{Studying at the library } | \text{ Being a business major}] = \frac{1}{2} = 0.50$[/tex]

[tex]$P(\alpha | B) = 0.50$[/tex]

From (1), we get

[tex]$P(\alpha | B^C) = \frac{1}{2} . P(\alpha | B)$[/tex]

              [tex]$=\frac{1}{2} \times 0.50$[/tex]

              = 0.25

Therefore, we need,

= P[  The students is a Business major | The student studies at the library ]

[tex]$=P(B | \alpha)$[/tex]

By Bayes theorem

[tex]$=\frac{P(B). P(\alpha | B)}{P(B).P(\alpha | B) + P(B^C). P(\alpha | B^C)}$[/tex]

[tex]$=\frac{0.45 \times 0.50}{0.45 \times 0.50 + 0.55 \times 0.25}$[/tex]

= 0.6207

Answer:

[tex]\displaystyle x \approx -4.28[/tex]

General Formulas and Concepts:

Pre-Algebra

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle 1 = ln(x + 7)[/tex]

Step 2: Solve for x

e^1 = x+7

e - 7 = x

x = -4.28