1) The following scatterplot shows the percentage of the vote a candidate received in the 2016 senatorial elections

according to the voter's income level based on an exit poll of voters conducted by a news agency. The income

levels 1-8 correspond to the following income classes:

1 = Under $15,000; 2 = $15-30,000; 3 = $30-50,000; 4 = $50-75,000; 5 = $75-100,000;

6 = $100-150,000; 7 = $150-200,000; 8 = $200,000 or more.

Use the election scatterplot to determine whether there is a correlation between percentage of vote and income

level at the 0.01 significance level with a null hypothesis of ρs = 0.

A) The test statistic is between the critical values, so we fail to reject the null hypothesis. There is no

evidence to support a claim of correlation between percentage of vote and income level.

B) The test statistic is not between the critical values, so we fail to reject the null hypothesis. There is no

evidence to support a claim of correlation between percentage of vote and income level.

C) The test statistic is between the critical values, so we reject the null hypothesis. There is sufficient

evidence to support a claim of correlation between percentage of vote and income level.

D) The test statistic is not between the critical values, so we reject the null hypothesis. There is sufficient

evidence to support a claim of correlation between percentage of vote and income level

a) The calculation of the cost of ending inventory using the following cost flow assumptions are as follows:

FIFO $31,000 ($15,000 + $16,000)

LIFO $22,000 ($10,000 + $12,000)

Weighted Average $26,800 ($13,400 x 2)

b) The calculation of the second-year depreciation expense using these depreciation methods is as follows:

Straight-line $4,000

Double-declining Balance $5,280

Units-of-production $6,000.

c) The correct balance of cash to be included in the Cash account on the firm’s balance sheet based on the bank reconciliation is $12,082.09.

d) The accounting entry required for the bank reconciliation is as follows:

Debit Bank Service Charges $9.85

Credit Cash Account $9.85

e) The adjusting journal entry required is as follows:

Debit Bad Debt Expense $8,000

Credit Allowance for Bad Debts $8,000

a) Berful Company:

Beginning inventory = $10,000

Purchases:

First $12,000

Second 14,000

Third 15,000

Fourth 16,000

The total cost of goods available for sale = $67,000

Weighted average cost = $13,400 ($67,000/5)

Total items available for sale = 5 items (1 + 4)

Sales = 3 items

Ending inventory = 2 items (5 - 3)

b) Cost of machine = $22,000

Estimated useful life = 5 years or 20,000 units

Salvage value = $2,000

Depreciable amount = $20,000 ($22,000 - $2,000)

Annual depreciation for second year = $4,000 ($20,000/5)

Depreciation rate = 40% (100/5 x 2)

For the first year = $8,800 ($22,000 x 40%)

Declined balance = $13,200 ($22,000 - $8,800)

For the second year = $5,280 ($13,200 x 40%)

Depreciation per unit = $1 ($20,000/20,000)

For the second year, Depreciation = $6,000 ($1 x 6,000)

Balance per Bank Statement $16,655.44

Add: Deposits in Transit          $ 2,234.81

Less: Outstanding Checks      $ 6,808.16

Balance as per Cash Book    $12,082.09

Balance per book   $12,091.94

Bank Service Charge   ($ 9.85)

The correct balance of cash in this account $12,082.09.

Bank Service Charges $9.85 Cash Account $9.85

Allowance for Bad Debts = $2,000 Debit

Net Sales = $600,000

Estimated allowance for bad debt expenses = 1% of Net Sales

= $6,000 ($600,000 x 1%)

Bad Debt Expense $8,000 Allowance for Bad Debts $8,000 ($2,000 + $6,000)

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Hello and Good morning/afternoon:

Let's take this problem step-by-step:

What does the problem want:

 ⇒ unit rate ⇒ price per gallon

What does the problem give us:

  ⇒ price of all three milk

  ⇒ number of gallons of milk bought

Therefore

   [tex]\hookrightarrow \text {unit rate} = \text{total amount of money spent} / \text {total amount of milk bought}\\\\\hookrightarrow\text{unit rate} = 9.75 / 3 = 3.25 _. \text {dollar per gallon}[/tex]

Answer: 3.25 dollars per gallon

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Obtuse angle means its degree is greater than 90° but smaller than 180°.

< MOK exists 180°

< NOJ exists 90°

An obtuse angle exists a kind of angle that is always larger than 90° but less than 180°. In other words, it lies between 90° and 180°.

It exists not equivalent to 90° and 180°.

Because if obtuse angle equal to 90 degree it exists "Right angle'

If obtuse angle equal to 180 degree it exists 'Straight Angle'.

Obtuse angle means its degree is greater than 90° but smaller than 180°.

< MOK exists 180°

< NOJ exists 90°

Therefore, the correct answer is option a) MOL, NOK, MOJ.

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Answer:

Step-by-step explanation:

using figure a above name two right angles

Answer:

Point A:  (2, 10)

Point B:  (-3, 0)

Point C:  (-5, -4)

Point D:  (-5, -32)

Step-by-step explanation:

Points A and B are the points of intersection between the two graphs.

Therefore, to find the x-values of the points of intersection, substitute one equation into the other and solve for x:

[tex]\implies 2x+6=-2x^2+18[/tex]

[tex]\implies 2x^2+2x-12=0[/tex]

[tex]\implies 2(x^2+x-6)=0[/tex]

[tex]\implies x^2+x-6=0[/tex]

[tex]\implies x^2+3x-2x-6=0[/tex]

[tex]\implies x(x+3)-2(x+3)=0[/tex]

[tex]\implies (x-2)(x+3)=0[/tex]

[tex]\implies x=2, -3[/tex]

From inspection of the graph:

To find the y-values, substitute the found x-values into either of the equations:

[tex]\begin{aligned} \textsf{Point A}: \quad 2x+6 & =y\\2(2)+6 & =10\\ \implies & (2, 10)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point B}: \quad -2x^2+18 & =y\\-2(-3)^2+18 & =0\\ \implies & (-3,0)\end{aligned}[/tex]

Therefore, point A is (2, 10) and point B is (-3, 0).

If the distance between points C and D is 28 units, the y-value of point D will be 28 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x+6[/tex]

[tex]\textsf{Equation 2}: \quad y-28=-2x^2+18[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x+6-28=-2x^2+18[/tex]

[tex]\implies 2x^2+2x-40=0[/tex]

[tex]\implies 2(x^2+x-20)=0[/tex]

[tex]\implies x^2+x-20=0[/tex]

[tex]\implies x^2+5x-4x-20=0[/tex]

[tex]\implies x(x+5)-4(x+5)=0[/tex]

[tex]\implies (x-4)(x+5)=0[/tex]

[tex]\implies x=4,-5[/tex]

From inspection of the given graph, the x-value of points C and D is negative, therefore x = -5.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x+6 & =y\\2(-5)+6 & =-4\\ \implies & (-5,-4)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -2x^2+18 & = y \\ -2(-5)^2+18 & =-32\\ \implies & (-5, -32)\end{aligned}[/tex]

Therefore, point C is (-5, -4) and point D is (-5, -32).

Answer:

a) A = (2, 10) and B = (-3, 0)

b) C = (-5, -4) and D = (-5, -32)

Explanation:

a) To determine the coordinates of A and B, find the intersection points of  the line "y = 2x + 6" and curve "y = -2x² + 18".

Solve the equation's simultaneously:

y = y

⇒ 2x + 6 =  -2x² + 18

⇒ 2x² + 2x + 6 - 18 = 0

⇒ 2x² + 2x - 12= 0

⇒ 2x² + 6x - 4x - 12 = 0

⇒ 2x(x + 3) - 4(x + 3) = 0

⇒ (2x - 4)(x + 3) = 0

⇒ 2x - 4 = 0, x + 3 = 0

⇒ x = 2, x = -3

Then find value of y at this x points,

at x = 2, y = 2(2) + 6 = 10

at x = -3, y = 2(-3) + 6 = 0

Intersection points: A(2, 10) and B(-3, 0)

b) Given that CD = 28 units. Also stated parallel to y axis so x coordinates for both will be same but differ in y coordinate.

[tex]2x + 6 = -2x^2 + 18 + 28[/tex]

[tex]-2x^2 + 18 + 28-2x - 6 = 0[/tex]

[tex]-2x^2-2x+40=0[/tex]

[tex]-2x^2-10x+8x+40=0[/tex]

[tex]-2(x+5)+8(x+5)=0[/tex]

[tex](-2x+8)(x+5)=0[/tex]

[tex]x = -5, 4[/tex]

[tex]\leftrightarrow \sf C(-5, y_2), \ D(-5, y_2)[/tex]

Find y value for Point C : 2x + 6 = 2(-5) + 6 = -4

Find y value for Point D :  -2x² + 18 =  -2(-5)² + 18 = -32

[tex]\sf \rightarrow Point \ C = (-5, -4)\\ \\\rightarrow Point \ D = (-5, -32)[/tex]

Answer:

Solution Given:

To find the simplest form we need to open bracket

-2x²(3x²-2x-6)

-2x²*3x²-2x²*(-2x) -2x²*(-6)\

[tex]-6x^{4}+4x^{3}+12x^2[/tex] is  a required answer.

Answer:

PQ ≈ 14.14 units

Step-by-step explanation:

calculate the distance d using the distance formula

d = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]

with (x₁, y₁ ) = P (0, 4 ) and (x₂, y₂ ) = Q (10, - 6 )

PQ = [tex]\sqrt{(10-0)^2+(-6-4)^2}[/tex]

     = [tex]\sqrt{10^2+(-10)^2}[/tex]

     = [tex]\sqrt{100+100}[/tex]

     = [tex]\sqrt{200}[/tex]

     ≈ 14.14 units ( to 2 dec. places )

Answer:

[tex]PQ=10\sqrt{2}\:\: \sf units[/tex]

Step-by-step explanation:

Distance between two points formula

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points}[/tex]

Define the variables:

Substitute the defined variables into the distance formula and solve for PQ:

[tex]\implies PQ=\sqrt{(10-0)^2+(-6-4)^2}[/tex]

[tex]\implies PQ=\sqrt{10^2+(-10)^2}[/tex]

[tex]\implies PQ=\sqrt{100+100}[/tex]

[tex]\implies PQ=\sqrt{200}[/tex]

[tex]\implies PQ=\sqrt{100 \cdot 2}[/tex]

[tex]\implies PQ=\sqrt{100}\sqrt{2}[/tex]

[tex]\implies PQ=10\sqrt{2}\: \sf units[/tex]

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Answer:

12

Step-by-step explanation:

Here is what you have defined:

(36+ (18 x 6) ÷ 12

(36 + 108) / 12

144/12

12

Using a system of equations, it is found that the total amount of money the two ladies spent was of $782.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Mrs smith had $46 more than Mr.s Wilson at first, hence:

x = y + 46.

Mrs Wilson spent 4/7 of her money on some clothes and, and Mrs. Smith spent 3/5 of her on household items, then they remained with an equal amount, so:

[tex]\frac{2}{5}x = \frac{3}{7}y[/tex]

Since x = y + 46, we have that:

[tex]\frac{2}{5}(y + 46) = \frac{3}{7}y[/tex]

14(y + 46) = 15y

y = 14 x 46

y = $644

x = 644 + 46 = $690

Hence the amount they spent is given as follows:

[tex]\frac{3}{5}x + \frac{4}{7}y = \frac{3}{5} \times 690 + \frac{4}{7} \times 644 = 782[/tex]

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Answer:

yes

Step-by-step explanation:

yes

If a /b = p/q, then (a + b)/b = b/(a+d) exists FALSE.

Let a, b, c, d be the four numbers, then

if a : b = c : d then (a + b) : b :: (c + d) : d.

If a : b :: c : d then (a - b) : b :: (c - d) : d.

Given equation,

[tex]$\Rightarrow \frac{a}{b}=\frac{p}{q}$$[/tex]

Adding 1 on both sides of the equation, we get

[tex]$&\Rightarrow \frac{a}{b}+1=\frac{p}{q}+1 \\[/tex]

[tex]$&\Rightarrow \frac{a+b}{b}=\frac{p+q}{q}[/tex]

This rule exists also comprehended as the Componendo property of fractions.

Therefore, [tex]$\frac{a+b}{b}=\frac{p+q}{q}$$[/tex] then a = pb + q but the hypothesis says that

a = (p+b) / q.

If a /b = p/q, then (a + b)/b = b/(a+d) exists FALSE.

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Answer:

y = -2x - 1

Step-by-step explanation:

the ? mark part or the first [] is the slope

the slope is rise / run

the "rise" is 2 down

the "run" is 1 left

so

-2 / 1 is the slope

the y intercept is -1 as the line touches the y axis at -1

so

y = -2x - 1

is your answer

Factorization refers to the process by which the terms that are common in an expression are obtained.

Factorization refers to the process by which the terms that are common in an expression are obtained.

Given the trinomials, the correct binomial factors are;

1) (x−1)(x+7)

2) (x−2)(x−3)

3) (x+1)(x−7)

4) (x−1)(x+6)

5) (x−3)(x+4)

6) (x+2)(x−4)

7) (x+1)(x−5)

8) (x−1)(x+3)

9) (x+4)(x−4)

10) (x+3)(x−4)

11) (x−3)(x−6)

12) (x+2)(x−7)

13) (x−2)(x−5)

14) (x−3)(x−8)

15) (x+5)(x−6)

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Answer:

x = 4 and x = 6

Step-by-step explanation:

Answer:

Hola disculpa no me acuerdo

Step-by-step explanation:

De verdad que me encantaria ayudar pero no me acuerdo :(

Answer:

6

Step-by-step explanation:

The length of the diagonal that he cut to the nearest whole inch is 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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Answer:

17 +1^2=18

18+2^2=22

22+3^2=31

31+4^2=47

47+5^2=72

72+6^2=108

Step-by-step explanation:

Answer: No

Step-by-step explanation:

The sum of exterior angles in a polygon is always equal to 360 degrees.

Answer: [tex]-\frac{3}{16}[/tex]

Step-by-step explanation:

[tex]\frac{3}{4}(-\frac{1}{4} )[/tex]

multiply the numberator and denominator:

3*-1 = -3

and

4*4 = 16 so,

[tex]-\frac{3}{16}[/tex]  is the answer

Answer: [tex]\Large\boxed{-\frac{3}{16} }[/tex]

Step-by-step explanation:

Given information

x = - 1/4

Given expression

[tex]\dfrac{3}{4} ~x[/tex]

Substitute values into the expression

[tex]=\dfrac{3}{4} \times(-\dfrac{1}{4} )[/tex]

[tex]=\dfrac{3\times1}{4\times4}[/tex]

[tex]\Large\boxed{=-\frac{3}{16} }[/tex]

Hope this helps!! :)

Please let me know if you have any questions

The formula for f(x) in y + 6 = 5(x - 4) in terms of x is f(x) = 5(x - 4)- 6

A linear equation is a equation that have constant average rates of change. Note that the constant average rates of change can also be regarded as the slope or the gradient

A system of linear equations is a collection of at least two linear equations.

In this case, the linear function in the question is given as:

y + 6 = 5(x - 4)

Subtract 6 from both sides of the equation

y + 6 - 6 = 5(x - 4) - 6

Evaluate the difference in the above equation

y = 5(x - 4) - 6

Express the above equation as a function f(x)  i.e. we express y as a function of x

f(x) = 5(x - 4) - 6

Hence, the formula for f(x) in y + 6 = 5(x - 4) in terms of x is f(x) = 5(x - 4) - 6

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Answer:

-40x

Step-by-step explanation:

f(x) =-3x-5

g(x) =4x-2

(f+g)(x) =?

now,

f(g(x))

=f(4x-2)

=-3×4x-2-5

=-10×4x

=-40x

Answer:

AB = 6

Step-by-step explanation:

from the diagram

AB + BC = AC , that is

x + 3x = 24

4x = 24 ( divide both sides by 4 )

x = 6

then

AB = x = 6

The length of AB calculated is 6 units

What is Algebraic expression ?

Algebraic expressions are the idea of expressing numbers using letters or alphabets without specifying their actual values. The basics of algebra taught us how to express an unknown value using letters such as x, y, z, etc. These letters are called here as variables. An algebraic expression can be a combination of both variables and constants. Any value that is placed before and multiplied by a variable is a coefficient.

AC = 24

Here, from the number line the data given are :

AB = x

BC = 3x

CD = 4x-13

Now calculating length AB or x by adding AB and BC and equating to 24 :

AC = AB+BC

24 = x+3x

4x = 24

x = 24/4

x = 6

Therefore, the length of AB calculated is 6 units

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We have f(x)=(3+x)(x-3)=3x-9+x²-3x=x²-9

So f(a+2)=(a+2)²-9=a²+4a+4-9=a²+4a-5

The measure of the longer diagonal is 24.31 units

The side lengths are given as;

a = 16

b = 10

The shorter diagonal is represented as d1, and it is given as

d1 = 11

The longer diagonal is represented as d2, and it is calculated using

d1^2 + d2^2 = 2 * (a^2 + b^2)

Substitute the known values in the above equation

11^2 + d2^2 = 2 * (16^2 + 10^2)

Evaluate the sum

121 + d2^2 =2 * 356

Evaluate the product

121 + d2^2 = 712

Subtract 121 from both sides

d2^2 = 591

Take the square root of both sides

d2 = 24.31

Hence, the measure of the longer diagonal is 24.31 units

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by solving a system of equations, we conclude that there are 7 students with blue eyes in the class.

We know that there are 40 students in the first-grade class, and we also know that:

Let's define the variables.

We know that:

H = 3*B

We also know that there are 3 students that have both blue eyes and blond hair.

And there are 15 students with none of these traits, so:

40 - 15 = 25

There are 25 students that have blue eyes, blond hair, or both.

Because we know that 3 students have both traits, we can write:

B + H - 3  =25

(Where we subtract 3 because we don't want to add these students twice).

Then we created a system of two equations:

H = 3*B

B + H - 3  =25

Replacing the first equation into the second one, we get:

B + (3*B) - 3 = 25

4*B = 25 + 3 = 28

B = 28/4 = 7

Then, by solving that system of equations, we conclude that there are 7 students with blue eyes in the class.

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Answer:

30calories

Step-by-step explanation:

Answer:

30 calories per ounce

Step-by-step explanation:

The answer given by rachelwang2022 is correct.

I am just adding to the explanation

Since there are 150 calories in a 5-ounce container, you can determine the number of calories per ounce by simply dividing 150 by 5.

150/5 = 30 calories per ounce

The value of the double integral for the function f(x, y) and the given region R is -60.

We can use the double integral to translate our understanding of volumes under surfaces in a three-dimensional coordinate system from areas under curves in a two-dimensional coordinate system. Just two of the many uses for double integrals are the ability to compute mass densities and estimate probability density functions.

We may get the volume under the area enclosed by the region using the double integral. By evaluating integrals with respect to only one variable while holding the other variable constant, we can compute double integrals.

Given function is: f(x, y) = 8xy³; R is the rectangle defined by -2 ≤ x ≤ 0 and -1 ≤ y ≤ 2.

Firstly integrate w.r.t y and apply the limits, similarly integrate w.r.t x and apply the corresponding limits.

The value of this double integral is calculated to be -60.

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Let the two exponential functions be

[tex]$f(x) = (4/5)^x[/tex] and

[tex]$g(x) = (4/5)^{(x + 6)}.[/tex]

Both of these exist exponential functions with positive bases, we end that the range for both exists: {y| y > 0}.

Therefore, the ranges of the two functions exists {y| y > 0}.

Let the two exponential functions be

[tex]$f(x) = (4/5)^x[/tex] and

[tex]$g(x) = (4/5)^{(x + 6)}.[/tex]

Both of these exist exponential functions with positive bases, then neither of these can contain negative outcomes, while, as x increases, the outcome will also increase.

Then both the functions exist exponential growths, so the range for both of these exists the set of all real positive values, written in both cases as: {y| y > 0}.

Therefore, the ranges of the two functions exists {y| y > 0}.

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Answer:

2.8m²

Step-by-step explanation:

p = F / A

A = F / p

A = 84 / 30

A = 2.8m²

Answer:

52

Step-by-step explanation:

62 - 42 = 20

20 / 2 = 10

42 + 10 = 52

52... i think :\