The average time between accidents in a factory is 5 weeks.
Find the probability that more than 7 weeks pass between accidents.

Answer:

This point is not on the line, NOT TRUE

Step-by-step explanation:

The equation of the line is written in the slope-intercept form,

y = mx + b

where m represents the slope

           b represents the y-intercept

where the format of a point is (x, y)

Given: y = 1/4x + 5, (−1, 6)

we can plug in the point (−1, 6)

6 = 1/4(-1) + 5

6 = -1/4 + 5

6 = 4.75

6 does not equal 4.75 so this point is not on the line

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The probability that exactly 6 of 8 adult use their smartphones in meetings or classes is 0.147 or 14.7% .

The complete question is

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an LG Smartphone survey). If 8 adult smartphone users are randomly selected, find the probability that exactly 6 of them use their smartphones in meetings or classes.

Probability is the measure of likeliness of an event to happen.

The probability given is

p = 54% = 0.54

n =8

r = 6

The binomial Probability is given by

[tex]\rm P(x) = ^n C_r p^r q^{n-r}[/tex]

The probability that exactly 6 people use smartphones in a class of 8 student is given by

[tex]P(x) = \dfrac{n!}{r! (n-r)!} p^r q^{n-r}[/tex]

Here q represents the complement

q = 1 -p

q = 1-0.54 = 0.46

[tex]P(6) = \dfrac{8!}{6! (2)!} 0.54^6 0.46^{2}\\\\\P(6) = \dfrac{8 * 7}{ 2} 0.54^6 0.46^{2}\\\P(6) = 0.147[/tex]

Therefore The probability that exactly 6 of 8 adult use their smartphones  in meetings or classes is 0.147 or 14.7% .

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Answer:

(x-6y)(x-4y)

Step-by-step explanation:

Answer:

[tex]-2x^2-x[/tex]. Your answer was wrong because "x-squared" terms didn't cancel.

Step-by-step explanation:

To solve this problem, set up an equation.  We know something is supposed to be added to the expression [tex]2x^2+2x[/tex], and the result should be x. So:

[tex]2x^2+2x+(\text{ ? })=x[/tex]

We want to solve for the question mark... the unknown thing that we're adding to the original expression, in order to get x.

It is uncommon to put question marks in equations to represent quantities. Usually we use a letter. Since x is already being used in the equation, we should pick something else ... we could use "y".

[tex]2x^2+2x+(\text{ } y \text{ })=x[/tex]

...or just...

[tex]2x^2+2x+y=x[/tex]

Algebra allows us to solve the equation and find out what "y" is equivalent to.

To solve, we want to get the "y" by itself. To do so, we try to eliminate the other "terms" from the left side of the equation.

Understanding "terms" & "like terms"

Terms

"Terms" in an equation are either a number multiplied to other things, or just a single number that isn't multiplied to anything else.

For example, the various terms in our equation above are

[tex]2x^2[/tex], [tex]2x[/tex], [tex]y[/tex], [tex]x[/tex]

You might ask why the last things, which don't have a number, are considered terms.

Remember that multiplying by 1 doesn't change anything, so we could imagine each of the last two terms as being 1 times the letter.

So, we can rewrite our equation:

 [tex]2x^2+2x+1y=1x[/tex]

Like terms

"Like terms" are terms where the "other stuff the numbers are multiplied to" is the same, so for instance, the [tex]2x[/tex] and the [tex]1x[/tex] are like terms. They are like terms because, the "other stuff" that the numbers are multiplied to are "x" for both terms. Note that [tex]2x[/tex] and [tex]2x^2[/tex] are not "like terms" because the "stuff" is different:

[tex]x[/tex] is different than [tex]x^2[/tex]

"Like terms" are important because only like terms can be "combined" into a single simplified term.

Solving equations

To solve an equation, we isolate what we're solving for, y, by disconnecting the other terms from it, and simplify.

Starting with subtracting 2x from both sides of the equation:

[tex]2x^2+2x+1y=1x\\(2x^2+2x+1y)-2x=(1x)-2x[/tex]

Subtraction is the same as "adding a negative":

[tex]2x^2+2x+1y+(-2x)=1x+(-2x)[/tex]

Since all terms are now connected by addition, we can add in any order we want (because of the Commutative Property of Addition), and we can combine like terms.

Thinking just about the number parts, since [tex]1+(-2)=-1[/tex],  then [tex]1x+(-2x)=-1x[/tex].

Returning to our main equation, the right side simplifies:

[tex]2x^2+2x+1y+(-2x)=1x+(-2x)\\2x^2+2x+1y+(-2x)=-1x[/tex]

On the left side: [tex]2x[/tex] and [tex]-2x[/tex] are like terms.

Fact: [tex]2+(-2)=0[/tex]

So, [tex]2x+(-2x)=0x[/tex]

Since anything times zero is just zero, [tex]0x=0[/tex]. Furthermore, adding zero to anything doesn't change it.  So when the [tex]2x[/tex] and [tex]-2x[/tex] terms on the left side of our main equation are combined, they "disappear" (While we talked through are a lot of rules/steps to justify why that works, it is common to omit those justifications, and to just combine those like terms and make them disappear.)

So, [tex]2x^2+2x+1y+(-2x)=-1x[/tex] simplifies to:

[tex]2x^2+1y=-1x[/tex]

Similarly for the [tex]2x^2[/tex] term, we subtract from both sides:

[tex]2x^2+1y=-1x\\(2x^2+1y)-2x^2=(-1x)-2x^2\\2x^2+1y+(-2x^2)=-1x+(-2x^2)[/tex]

Combining like terms on the left, they disappear.

[tex]1y=-1x+(-2x^2)[/tex]

There are no like terms on the right.

Since the two terms on the right are added together, we can use the commutative property of addition to rearrange:

[tex]1y=-2x^2+(-1x)[/tex]

Addition of a negative can turn back into subtraction, and simplify multiplication by 1.

[tex]y=-2x^2-x[/tex]

Remembering we chose "y" as the unknown thing we wanted to know, that's why the "correct answer" is what it is.

Verifying an answer

Verifying can double check an answer, and helps explain why the answer you chose doesn't work.

To verify an answer, the original statement said add something to the expression and get a result of "x". So, let's see if the "correct answer" does:

[tex]2x^2+2x+(\text{ } ? \text{ })\\2x^2+2x+(-2x^2-x)\\2x^2+2x+(-2x^2-1x)\\2x^2+2x+(-2x^2)+(-1x)[/tex]

Combining the "x-squared" terms, completely cancels...

[tex]2x+(-1x)[/tex]

Combining the "x" terms, and simplifying...

[tex]1x\\x[/tex]

So it works.

Why isn't the answer what you chose:

[tex]2x^2+2x+(\text{ } ? \text{ })\\2x^2+2x+(-x^2-x)\\2x^2+2x+(-1x^2-1x)\\2x^2+2x+(-1x^2)+(-1x)[/tex]

Combining the x-squared terms, things don't completely cancel...

[tex]1x^2+2x+(-1x)[/tex]

Combining the x terms...

[tex]1x^2+1x\\x^2+x[/tex]

So adding the answer that you chose to the expression would not give a result of "x", which is why it is "wrong"

Answer:

  (b)  3 terms, degree 10

Step-by-step explanation:

When simplifying polynomials, the first thing you look at is the exponents of the variables in each term. (The terms are separated by + or - signs.) Only terms with the same set of exponents can be combined.

When considering the degree of the polynomial, you look at the sum of exponents of the variables in each term. That sum is the degree of the term. The degree of the polynomial is the highest of the degrees of the terms.

__

The four terms have variables s and t. If we write the exponents of those variables as an ordered pair, we will have one ordered pair for each term:

  5s⁶t²  ⇒  (exponent of s, exponent of t) ⇒ (6, 2); degree = 6+2 = 8

  6st⁹  ⇒  (1, 9); degree = 1+9 = 10

  -8s⁶t²  ⇒  (6, 2); degree 6+2 = 8 (this is a like term to the first term)

  -6t⁷  ⇒  (0, 7); degree 0+7 = 7

The degrees are 8, 10, 8, 7, so the highest degree is 10. This is the degree of the polynomial.

__

We can only combine the terms containing s⁶t². Doing that gives ...

  (5 -8)s⁶t² +6st⁹ -6t⁷ = -3s⁶t² +6st⁹ -6t⁷ . . . . . three terms

In "standard form", the terms are written in order of decreasing degree:

  6st⁹ -3s⁶t² -6t⁷ . . . . . . 3 terms, degree 10

Answer: 12.4 x 18

-------------- = 18600

0.012

The probability of picking a hazelnut chocolate is 1/5 and the number of hazelnut chocolates in the box is 1

Note that the ratios =  5:4:2:3

toffee: coffee: orange: mint chocolates = 5:4:2:3:1

Total ratio = 15

Probability of picking a hazelnut chocolate = ratio of hazelnut chocolates/ total ratio

Probability of picking a hazelnut chocolate  = [tex]\frac{1}{15}[/tex]

The number of hazelnut chocolates in the box is gotten from the ratio which is 1

Therefore, the probability of picking a hazelnut chocolate is 1/5 and the number of hazelnut chocolates in the box is 1

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Answer: I hope this helps you in what you're doing.

Step-by-step explanation: Because multiplying two rational numbers results in the ratio of two integers, which is also a rational number, the product of two rational numbers is also a rational number.

The e vertex of the decagon will be in the top position after rotating it counterclockwise by 3 times the smallest angle of rotation.

Which vertex will be in the top position of the regular decagon?

A regular decagon has 10 sides of equal lengths with points labeled 'a' through 'j' clockwise. It is given that the point a is the top-left point. Hence, the the vertex which is in the the top position currently is 'b'.

Now, the smallest angle of rotation will be the angle between the two sides of the decagon.

In the first rotation by the smallest angle in counterclockwise direction, point 'c' will come to the top position. In the second rotation by the smallest angle in counterclockwise direction,  point 'd' point will become the top most vertex. Finally, after the third similar rotation, 'e' vertex will be in the top position of the decagon. (Refer the attached diagram)

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well if its asking for an exponent then its (-5)⁴

Answer:

3

Step-by-step explanation:

Answer:

The y-intercept is 22 spaces higher.

Step-by-step explanation:

Answer:

The value of x is 117°  by using Alternative interior angle theorem

Step-by-step explanation:

Alternative interior angle theorem: It states that the angles made by transverse between two parallel lines are congruent.

as in figure

Angle A = 42°

Angle B = 108°

Angle C = x

Angle D = 51°

draw two parallel lines passing through B and C .

According to Alternative interior angle theorem angle B is written as 42° + 66°

and angle C is written as x1 + x2 .

Now x1 = 51° as x1 and D are alternative angles

And similarly x2 = 66°

As x = x1 + x2

Therefore x = 51° + 66° = 117°

The value of x is 117°

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[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex] \tt{sin(\Theta) = \cfrac{3}{8}} [/tex]

[tex]\qquad \tt \rightarrow \: x = 6\:\: cm[/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: \sin( Theta) = \cfrac{Opposite \:\: side}{Hypotenuse}[/tex]

[tex]\qquad \tt \rightarrow \: \sin( \Theta) = \cfrac{3}{8} [/tex]

since the Triangles are similar, ratio of their corresponding sides are equal :

[tex]\qquad \tt \rightarrow \: \sin( \Theta) = \cfrac{3}{8} = \dfrac{x}{16} [/tex]

[tex]\qquad \tt \rightarrow \:x = 16 \sdot \cfrac{3}{8}[/tex]

[tex]\qquad \tt \rightarrow \: x = 2 \sdot3[/tex]

[tex]\qquad \tt \rightarrow \: x = 6 \: \: cm[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Since  the two right-angled triangles have the same angles, the values are; ai. sin θ= 3/8ii. sin θ= x/16b. x = 6cm

Using the sine trigonometric identity expressed as;

sin θ = opposite/hypotenuse

Then, we have that for the first triangle;

sin θ= 3/8

For the bigger triangle;

sin θ= x/16b.

Then, since we have that the sine identity is;

sin θ= x/16

But sin θ= 3/8

Divide the values and find the sine inverse in

θ= 0. 375θ = 22. 0 degrees

x = 0. 375 (16)

Multiply

x = 6 cm

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Step-by-step explanation:

Well the first 3 you just plug in the value of x which results in.

a. f(0) = 7(0)^2 + 3(0) - 2

   f(0) = -2

b. f(2) = 7(2)^2 + 3(2) - 2

   f(2) = 7(4) + 6 - 2

   f(2) = 28 + 4

   f(2) = 32

c. f(-2) = 7(-2)^2 + 3(-2) - 2

   f(-2) = 7(4) - 6 - 2

   f(-2) = 28 - 8

   f(-2) = 20

for the last ones it's essentially the same thing except the value a is not known

d. f(a+1) = 7(a+1)^2 + 3(a+1) - 2

   f(a+1) = 7(a^2 + 2a + 1) + 3a + 3 - 2

   f(a+1) = 7a^2 +14a + 7 + 3a + 1

   f(a+1) = 7a^2 + 17a + 8

c. f(-a) = 7(-a)^2 + 3(a) - 2

   f(-a) = 7a^2 + 3a - 2

The function over the interval is an increasing exponential function.

A quadratic function is" a polynomial function with one or more variables in which highest exponent of variable is 2".

The complete question is

An investment group compares returns on an account

against the function represented in the table, where x is the

time in years and f(x) is the total return on investment.

Which describes the function over the interval given in the

table?

х

a decreasing quadratic function

an increasing quadratic function

a decreasing exponential function

an increasing exponential function

0

5

f(x)

10,000

12,201.90

14,888.64

22,167.15

10

20

A decreasing quadratic function is the vertex of the parabola lies on the axis parabola. So, time in years change does not give maximum total return on investment.

For increasing quadratic equation, the graph increasing at one side of the axis and decreases at other side.

For, Exponential function          

When the exponential function then it shows total return on investment is not maximum.

When the exponential function is increasing it shows time goes, total return on investment is maximum.

Hence, the exponential function is increasing.

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C- 1/5

Step-by-step explanation:

a = 2 - 9a

10a = 2

a = 2/10

a = 1/5

Hope u got it (◠‿・)—☆

Answer:

the answer you're looking for is 60 inches

The perimeter of the base is 4 s since it is a square.

The area of the base is s².

[tex]\boldsymbol{CST=\dfrac{1}{2}(64)(17)+256=544+256=800 \ inches^{2} }[/tex]

[tex]\red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}[/tex]

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:Domain = [-9, -1][/tex]

[tex]\qquad \tt \rightarrow \:Range = [-1 , 3][/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

Domain = All possible values of x for which f(x) is defined

[ generally the extension of function in x - direction ]

Range = All possible values of f(x)

[ generally the extension of function in y - direction ]

[tex] \large\textsf{For the given graph : } [/tex]

[tex]\qquad \tt \rightarrow \: domain = [ - 9, -1][/tex]

[tex]\qquad \tt \rightarrow \: range= [ -1,3][/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

The value of the element in row 1, column 1 of matrix x is; y.

According to the task content, it follows that matrix x is defined as; x= [y 4].

In essence, it follows that the matrix x contains one row and 2 columns. Therefore, the element in row 1, column 1 of matrix x is; y.

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Answer: A

Step-by-step explanation:

Answer:

20 + 4y

Step-by-step explanation:

To have to fewest symbols possible, we must simplify the expression by multiplying 4 into the numbers within the parenthesis. Doing so we get:

[tex]4(5+y) = 4 * 5 + 4 * y = 20 + 4y[/tex]

Answer: 20+4y

Step-by-step explanation:

Hi, let me help you.

What we should do to simplify this expression is to "distribute" 4.

The parentheses around 5+y tell us to multiply 4 times both 5 and y. This is shown below.

[tex]\multimap\quad \tt 4\cdot5+4\cdot y[/tex]

[tex]\multimap\quad\tt 20+4y[/tex] is what we obtain upon multiplying

________________

Hope I helped. Best wishes.

Reach far. Aim high. Dream big.

____________________

Answer:

Symmetric property

Answer:

40 miles

Step-by-step explanation:

10X4

The answer is 40 miles

-------------------------------------------------------

If he consistently rides ten mph (miles per hour) for 4 hours (4*10) he would have ridden 40 miles.

Answer: $0.24

Step-by-step explanation:

9.5 inches --> $2.28
1 inch--->X
now cross multiply it and get,

9.5X = 2.28*1
X = 2.28/9.5

so its $0.24 per inch

Answer:

[tex]\dfrac{26}{3}[/tex]

Step-by-step explanation:

The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:

[tex]\dfrac{f(b)-f(a)}{b-a}[/tex]

Given function:  [tex]f(x)=3^{x-1}+2[/tex]

Given interval:  1 ≤ x ≤ 4

Therefore, a = 1 and b = 4

Therefore, find the value of the given function when x = 1 and x = 4:

[tex]\begin{aligned}\implies f(1) & =3^{1-1}+2\\ & = 3^0+2\\ & = 1+2\\ & = 3 \end{aligned}[/tex]

[tex]\begin{aligned} \implies f(4) & = 3^{4-1}+2 \\ & = 3^3+2\\ & = 27+2\\ & = 29 \end{aligned}[/tex]

Substitute the found values into the formula:

[tex]\begin{aligned}\implies \dfrac{f(b)-f(a)}{b-a} & =\dfrac{f(4)-f(1)}{4-1}\\\\ & =\dfrac{29-3}{4-1}\\\\ & = \dfrac{26}{3} \end{aligned}[/tex]

    Sin and cos are trigonometry functions used to find measurements in a right triangle.

         [tex]\displaystyle sin=\frac{\text{opposite side}}{hypotenuse}[/tex]

         [tex]\displaystyle cos=\frac{\text{adjacent side}}{hypotenuse}[/tex]

The first difference and second difference of the equation will be given below.

The quadratic equation is given as ax² + bx + c = 0. Then the degree of the equation will be 2. Then we have

The quadratic function is given below.

y = x² + 7x + 6

Let the value of the y for x = 0, 1, 2, 3, 4 will be

y(0) = 6

y(1) = 14

y(2) = 24

y(3) = 36

y(4) = 50

The first difference and second difference will be

x       y      first difference        second difference

0       6               ...                                   ...

1       14              8                                   ...

2       24            10                                   2

3       36            12                                   2

4       50            14                                   2

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The exact value of [tex]\dfrac{cos(tan^{-1}(-1))}{sin(cos^{-1}(-\frac{3}{\sqrt{2}})}[/tex] would be root 2.

The function is a type of relation, or rule, that maps one input to specific single output.

Given;

[tex]\dfrac{cos(tan^{-1}(-1))}{sin(cos^{-1}(-\frac{3}{\sqrt{2}})}[/tex]

We know that

The inverse tan(-1) = -π/4

cos(-π/4) = 1/√2

The inverse cos(√3/2) = π//6

sin(π/6) = 1/2

Then substitute

[tex]\dfrac{cos(tan^{-1}(-1))}{sin(cos^{-1}(-\frac{3}{\sqrt{2}})}\\\\[/tex]

[tex]\dfrac{(1/\sqrt{2}) }{(1/2)} \\\\= \sqrt{2}[/tex]

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Answer:

a

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

here m = [tex]\frac{1}{3}[/tex] , then

y = [tex]\frac{1}{3}[/tex] x + c ← is the partial equation

to find c substitute (6, 4 ) into the partial equation

4 = 2 + c ⇒ c = 4 - 2 = 2

y = [tex]\frac{1}{3}[/tex] x + 2 ← equation of line