A spinner is divided into equal sections of three possible colors. There are 2 yellow
sections, 2 blue sections, and 4 red sections. Use the drop-down menus to complete the
statements about the likelihood of the given events related to the spinner.

Answer:

1/8

Step-by-step explanation:

I've done this before.

The cash advance fee for the credit card account is $3.00

What is the basis for computing cash advance fee?

The basis for determining the cash advance fee is on the amount borrowed not that 2% is charged annually on the credit card account, in other words, the cash advance fee in this case is the cash advance multiplied by the advance fee of 2% as shown explicitly below:

cash advance fee=cash advance*cash advance fee %

cash advance=$150

cash advance %=2%

cash advance fee=$150*2%

cash advance fee=$3.00

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Full question:

Sidney took a $150 cash advance by using checks linked to her credit card account. The bank charges a 2 percent cash advance fee on the amount borrowed and offers no grace period on cash advances. Sidney paid the balance in full when the bill arrived.

What was the cash advance fee? (Round your answer to 2 decimal places.)

Answer:

-1

Step-by-step explanation:

y = mx + c

m is the slope/gradient

to find slope = y2 - y1 / x2 - x1

= -8 - -10 / 4 - 6

m = -1

Answer:

following is the answer for your question this can be done in the manner

Answer:

24964

Step-by-step explanation:

hope this works : ) - btw this is not including 2 and 33

Answer:

Let the numbers be a, b, c, d

[tex] { \tt{a + b + c + d = 93}}[/tex]

Let a be the multiple of 19, a = 57

And b = c = d

[tex]{ \tt{57+ 3b= 93}} \\ { \tt{3b = 36}} \\ { \tt{b = 12}}[/tex]

[tex]{ \tt{a = 57}} \\ { \tt{b = c = d = 12}} \\ \\ { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{57}}}[/tex]

Answer:

12, 12, 12, 57

Step-by-step explanation:

Given:

If one of the four numbers is a multiple of 19 and the other three numbers are equal, then the sum of the other three numbers will be a multiple of 3.

Multiples of 19:  

19, 38, 57, 76, 95, ...

Subtract each multiple of 19 (that is less than 93) from 93:

⇒ 93 - 19 = 74  →  not divisible by 3

⇒ 93 - 38 = 55  →  not divisible by 3

⇒ 93 - 57 = 36  →  divisible by 3

⇒ 93 - 76 = 17  →  not divisible by 3

The only result that is a multiple of 3 is 36.  

Therefore, the multiple of 19 is 57.

To find the other three numbers, simply divide the result of the total less the found multiple of 19 by 3:

⇒ (93 - 57) ÷ 3 = 36 ÷ 3 = 12

Therefore, the four numbers in ascending order are:

12, 12, 12, 57

Answer:

The first option

Step-by-step explanation:

Answer:

1st one

Step-by-step explanation:

Opposite of positive 3 is negative 3.

Answer:

goood luccccccckkkk...

Answer:

Ratio: [tex]=\mathbf{\frac{6}{11}}[/tex]

The decimal form (showing up to four  decimal places): 0.5454

The decimal is a repeating decimal i.e 0.54545454..

Step-by-step explanation:

We need to find ratio of hockey players to soccer players

Number of players playing Hockey= 6

Number of players playing Soccer = 11

Ratio: [tex]\frac{Hockey\:Players}{Soccer\:Players} =\mathbf{\frac{6}{11}}[/tex]

The decimal form (showing up to four  decimal places): 0.5454

Tell whether the decimal is a terminating or a repeating decimal.

The decimal is a repeating decimal i.e 0.54545454..

Answer:

116, 117, 118

Step-by-step explanation:

Consecutive numbers are numbers that come after each other, such as 1, 2 and 3.

Find the expressions for the integers:

Let the smallest integer be x.

2nd integer= x +1

3rd integer= x +1 +1= x +2

Form an equation:

x +x +1 +x +2= 351

Solve for x:

3x +3= 351

3x= 351 -3

3x= 348

x= 348 ÷3

x= 116

Find the other 2 integers:

2nd integer= 116 +1= 117

3rd integer= 117 +1= 118

Hence, the three consecutive integers are 116, 117 and 118.

Answer:

11/12

Step-by-step explanation:

15/12-4/12= 11/12

Answer:

11/12

Step-by-step explanation:

5/4 -1/3 first make the equation easier to look at so take the 1 and add it as 4/4

15/12-4/12 then make the bottom the same with multipulcation and tops are multiples as well

11/12 subtract

Answer:

it's c i did this a week ago

Answer:

B

Step-by-step explanation:

(look up limit calculator)

a has a limit

b does not exist

c  has a limit

d has a limit

e has a limit

Option D is correct one. Limit doesn't exist for [tex]\lim_{x\to \ 2} \frac{x^{2}-6x +8 }{x-2}[/tex].

The limit of f(x) at x = a is said to be exist if the function approaches the same value from both the sides i.e. LHL =RHL (left hand limit = right hand limit)

According to the question we have,

A. [tex]\lim_{x \to \ 1} \frac{2-x}{x+1}[/tex]

at x =1

LHL = [tex]\lim_{h \to \ 0 } \frac{2- (1-h)}{(1-h)+1}[/tex] = [tex]\frac{1}{2}[/tex]

RHL = [tex]\lim_{h \to \0} \frac{(2-(1+h) }{(1+h)+1)} =\frac{1}{2}[/tex]

Here, RHL = LHL

Hence, limit exist.

B. [tex]\lim_{x \to \ 2} \frac{x^{2} -8}{x-2}[/tex]

at x =2

LHL = [tex]\lim_{h \to \ 0} \frac{(2-h)^{2}-9 }{(2-h)-2}[/tex] =[tex]\lim_{h \to \ 0} \frac{4+h^{2} -9-4h}{2-h-2}[/tex] = [tex]\lim_{h \to \ 0 }\frac{2h-4}{-1} =\frac{-4}{-1} =4[/tex]

RHL = [tex]\lim_{h \to \ 0} \frac{(2+h)^{2} -9}{(2+h)-2}[/tex] = [tex]\lim_{h \to \ 0} \frac{4h +4 +h^{2}-9 }{h} = \lim_{h \to \ 0} \frac{2h+4}{1}[/tex]=4

Here, RHL = LHL

Hence, the limit exist.

C. [tex]\lim_{x \to \ 1} (x^{2} -2x-5)[/tex]

at x =1

LHL = [tex]\lim_{h \to \ 0} (( 1-h)^{2} -2(1-h)-5)= \lim_{h \to \ 0} (1 -2-5)=-6[/tex]

RHL = [tex]\lim_{h \to \ 0} ((1+h)^{2} -2(1+h)-5) = (1-2-5)=-6[/tex]

Here, LHL = RHL = -6

Hence, the limits exist.

D. [tex]\lim_{x \to \ 2} \frac{x^{2}-6x + 8 }{x - 2}[/tex]  

at x =2

LHL = [tex]\lim_{h \to \0} \frac{(2-h)^{2} -(2-h)+8}{(2-h)-2} = \lim_{h \to \ 0} \frac{4+h^{2}-4h-2+h+8 }{(2-h)-2}[/tex] =[tex]\lim_{h \to \ 0} \frac{2h-4+1}{1} =-3[/tex]

RHL = [tex]\lim_{h \to \ 0} \frac{(2+h)^{2} -(2+h) +8}{(2+h)-2} = \lim_{h \to \ 0} \frac{4+h^{2}+4h -2-h+8}{h} = \lim_{h \to \ 0} \frac{2h+4-1}{1} =3[/tex]

Here, RHL≠LHL

Hence, the limit doesn't exist.

E. [tex]\lim_{x \to \ 3} (x^{2} -3x-1)[/tex]

at x = 3

LHL = [tex]\lim_{h \to \ 0} ((3-h)^{2} -3(3-h)-1)[/tex] = [tex]9-9-1=-1[/tex]

RHL= [tex]\lim_{h \to \ 0} ((3+h)^{2} -3(3+h)-1)= 9-9-1 =-1[/tex]

Here, RHL = LHL

Hence, the limit exist.

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Answer:

7/8

Step-by-step explanation:

Using the binomial distribution, it is found that there is a [tex]\frac{7}{8}[/tex] probability that at least one of the coins comes up heads.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

In this problem:

The probability that at least one of the coins comes up heads is given by:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{3,0}.\left(\frac{1}{2}\right)^{0}.\left(\frac{1}{2}\right)^{3} = \frac{1}{8}[/tex]

Then:

[tex]P(X \geq 1) = 1 - P(X = 0) = \frac{8}{8} - \frac{1}{8} = \frac{7}{8}[/tex]

[tex]\frac{7}{8}[/tex] probability that at least one of the coins comes up heads.

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Answer:

The mass of an object is directly proportional to its inertia.

[tex]m\alpha I[/tex]

Where

m = mass

I = inertia

Step-by-step explanation:

We can first of all start by defining the two key terms present here - mass and inertia.

The mass of an object is the quantity of matter contained in the object.

The inertia of an object is the resistance of an object which is in a state of rest to motion

The mass of an object is directly proportional to its inertia.

The heavier an object is, the more inertia it has. To put things into perspective, you can easily push a bicycle because it does not really have a lot of mass, as opposed to pushing an eighteen-wheeler trailer.

Answer: yes they are equivalent

Step-by-step explanation: When you distribute both equations, you end up with 5m-10.

Answer:

9/26

Step-by-step explanation:

2/159÷2/477=2/53÷2/159=6/52÷2/53=3

x÷6/52=3

x=18/52

x=9/26

Answer:

Step-by-step explanation:

4^-1

Answer:

It is a function

Step-by-step explanation:

A function is of the form:

[tex]\left[\begin{array}{c}x_1&x_2&x_3\\-&-&-\\x_n\end{array}\right][/tex]            [tex]\left[\begin{array}{c}y_1&y_2&y_3\\-&-&-\\y_n\end{array}\right][/tex]

Where each of the x values must be distinct.

The x values are referred to as domain while the y values are called range.

Having said that, the given data can be represented as follows:

[tex]\left[\begin{array}{c}-3&1&6\\7&12\end{array}\right][/tex] ----------- [tex]\left[\begin{array}{c}7&-8&-6\\-8&2\end{array}\right][/tex]

From the representation above, none of the x values are repeated.

Hence, it is a function

Answer:

f(t) = -300t + 20,700

Step-by-step explanation:

2000 is the starting year, so 20,700 is our y-intercept or our constant.

Each year, represented by t, student enrollment decreases by 300. This is the slope of the equation.

Answer:

D

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

Answer:

Step-by-step explanation:

Part A: No. The line would have to be straight in order for the graph to be linear.

Part B:

Linear: y=2x-4.

Nonlinear: y=x^2-6.

Answer:  No.

Step-by-step explanation: Unless the tenth place is a 0, a product of a 2 numbers with 1 decimal place cannot have a product for a whole number, even if you went on forever.

When two decimal places add up to one whole, the total of the two decimal places is a whole number. The addition of relocating the decimal point after the multiplication when multiplying decimals is the only difference between multiplying whole numbers and multiplying decimals.

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Answer:

HAHAHAHH

Step-by-step explanation:

The segment joining the midpoints of the diagonals of a trapezoid is parallel to the bases

From the figure, we have the following coordinates

B = (a, 0)

D = (d, c)

The midpoint of line segment BD is calculated as

BD = 1/2 * (x1 + x2, y1 + y2)

So, we have

BD = 1/2 * (a + d, 0 + c)

This gives

BD = ((a + d)/2, c/2)

Also, we have:

Lines MN and AB are horizontal lines

This means that they have a slope of 0

i.e.

Slope of AB = Slope of MN = 0

Parallel lines have equal slope

Hence, the segment joining the midpoints of the diagonals of a trapezoid is parallel to the bases

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