Write an equation to represent this scenerio
You are about to take a major assessment, and each question is one two types. Multiple-choice questions are worth 2 points each, and short-response questions are worth 4 points each. The assessment is worth 100 total points. Let x represent the number of multiple-choice questions, and let y represent the number of short-response questions.

Answer:

-3x-5 x-4 I this work good luck

Answer:

-(3x-5) (x-4)

Step-by-step explanation:

-3x^2+17x-20

Factor -1 out of -3x^2

-(3x^2)+17x-20

Factor -1 out of 17x

-(3x^2)-(-17x)-20

Rewrite -20 as -1(20)

-(3x^2)-(-17x)-1(20)

Factor -1 out of -(3x^2)-(-17x)

-(3x^2-17x)-1(20)

Factor -1 out of -(3x^2-17x)-1(20)

-(3x^2-17x+20)

Factor.

-((3x-5)(x-4))

Final Anwser:

-(3x-5) (x-4)

Answer:

x=(1)/(3),-(3)/(2)

Step-by-step explanation:

Answer:

(- 3/2, 1/3)

Step-by-step explanation:

7x^2+10x-3=x^2+3x

7x^2+10x-3-x^2-3x=0

6x^2+7x-3=0

6x^2+9x-2x-3=0

3x x (2x+3)-(2x+3)=0

(2x+3)x(3x-1)=0

2x+3=0

3x-1=0

x1=- 3/2

x2= 1/3

Answer:

KL = 16

Step-by-step explanation:

For this problem, JL is the sum of KL and JK.  So we can say this:

JK + KL = JL

( 3x + 6 ) + ( 5x + 6 ) = 28

8x + 12 = 28

8x = 16

x = 2

So, now we can find KL:

5x + 6 = ?

5(2) + 6 = ?

10 + 6 = ?

16 = ?

So the length of KL is 16.

Cheers.

Answer:

173, 339.10

Step-by-step explanation:

Move the decimal from 157,581.00 to the left to give you 10% of 15,758.10.

Add 15,758.10 to 157,581.00 to give you the total from last month.

Or, convert 10% to a decimal.

157,781 x 0.10 = 15,758.10

15,758.10 + 157,581 = 173,339.10

The required value of this game in the long run has no value, as game will discarded because both players identical pairs.

A standard 52-card deck contains 13 positions in every one of the four French suits: clubs (♣), precious stones (♦), hearts (♥) and spades (♠). Each suit incorporates three court cards (face cards), Lord, Sovereign and Jack, with reversible (twofold headed) pictures. Each suit additionally incorporates ten numeral cards or pip cards, from one to ten.

After the blended matches are disposed of, there are an equivalent number of dark red still in the deck, so the two players will have similar number of pairs.

So, No one will win the game.

Thus, the game will  discarded.

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Answer:

The man paid $200 for the cow

Step-by-step explanation:

System of Equations

Let's call:

x = price of the cow

y = price of the horse

The man bought the cow and the horse for $500, thus

x + y = 500        [1]

The cow was sold at a profit of 10%, thus:

Sale price of the cow= 1.1x

The horse was sold at a loss of 10%, thus:

Sale price of the horse= 0.9y

The total operation was a 2% loss, i.e. 0.98*500=490. Thus, we have:

1.1x + 0.9y = 490        [2]

From [1]:

y = 500 - x

Substituting in [2]:

1.1x + 0.9(500 - x) = 490

Operating:

1.1x + 450 - 0.9x = 490

0.2x = 490 - 450 = 40

x = 40/0.2

x = 200

The man paid $200 for the cow

Step-by-step explanation:

yo question doesn't make sense.

Answer:   C

.......................

Answer: A

Step-by-step explanation:

The object represented by h(t) = -4.9t² + 40t + 20 would take 4.082 seconds to reach it's maximum height.

An equation is an expression that shows the relationship between two numbers and variables.

An independent variable is a variable that does not depend on any other variable for its value whereas a dependent variable is a variable that depend on any other variable for its value.

The height (h) is represented by:

h(t) = -4.9t² + 40t + 20

The maximum height is at h'(t) = 0, hence:

h'(t) = -9,8t + 40

-9.8t + 40 = 0

t =  4.082

The object represented by h(t) = -4.9t² + 40t + 20 would take 4.082 seconds to reach it's maximum height.

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Answer:

first second and third i believe

Step-by-step explanation:

Answer:

the first and third

Step-by-step explanation:

hope i was right lol

To calculate the cash balance at the end of the period, we need to consider the cash inflows and outflows.

Starting cash balance: $4,520

Cash inflows: $1,654 (receivables collected)

Cash outflows: $1,961 (payments to suppliers) + $500 (cash expenses)

Total cash inflows: $1,654

Total cash outflows: $1,961 + $500 = $2,461

To calculate the cash balance at the end of the period, we subtract the total cash outflows from the starting cash balance and add the total cash inflows:

Cash balance at the end of the period = Starting cash balance + Total cash inflows - Total cash outflows

Cash balance at the end of the period = $4,520 + $1,654 - $2,461

Cash balance at the end of the period = $4,520 - $807

Cash balance at the end of the period = $3,713

Therefore, the cash balance at the end of the period is $3,713.

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The results of the significance test suggest that there is no statistically significant change in the average consumption of sweet snacks among healthy weight children aged 12 to 19 years.

The previous study reported an average consumption of 369.4 kcal/d, with a standard deviation of 10.2 kcal/d. The current study reported an average consumption of 375.2 kcal/d, with a standard deviation of 10.2 kcal/d.

To perform a significance test, we can use the z-score formula:

z = (observed value - mean) / standard deviation

In this case, the observed value is 375.2 kcal/d, the mean is 369.4 kcal/d, and the standard deviation is 10.2 kcal/d. This gives us a z-score of 0.56.

The p-value for a z-score of 0.56 is 0.5696. This means that there is a 56.96% chance of getting a z-score of 0.56 or greater if the null hypothesis is true.

Since the p-value is greater than the significance level of 0.05, we cannot reject the null hypothesis. This means that there is not enough evidence to conclude that there has been a statistically significant change in the average consumption of sweet snacks.

In other words, the results of the current study are consistent with the results of the previous study. There is no evidence to suggest that the average consumption of sweet snacks has changed in the past four years.

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Answer:

C

Step-by-step explanation:

52+15*90-6029=19.2

No, the language L is not a deterministic context-free language (DCFL).

To see if L is deterministic, suppose L is a DCFL. Then there exists a deterministic pushdown automaton (DPDA) that recognizes L.

Consider the string w = a^p b^(p+1) c^(p+1) ∈ L, where p is the pumping length of L. Since w is in L and L is a DCFL, the DPDA for L must accept w.

Assuming that the DPDA for L has only one accepting state. Let q be this accepting state.

By the pigeonhole principle, Let u, v, and x be the three parts of w such that u and v are the substrings of w that correspond to the first two occurrences of q', and x is the remaining suffix of w.

Then we can pump v any number of times and still get a string in L.

We can make the number of b's divisible by the number of c's by choosing an appropriate number of pumps.

However, since v contains at least one b, pumping v will result in a string that contains more b's than c's, which is not in L.

Therefore, we have a contradiction, and L cannot be a deterministic context-free language.

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False, The eigenvalues of a matrix A are not same as the eigenvalues of the reduced row echelon form of A.

A matrix's eigenvalues are not the same as those of the matrix's reduced row echelon form, contrary to what is claimed.

This is due to the fact that when a matrix is reduced to its echelon form via row operations, its eigenvalues may not always stay the same. The eigenvalues of a matrix may be different in the reduced row echelon form than they are in the original matrix as a result.

Think about these matrices as an illustration:

Likewise, B's reduced row echelon form is B, and B's eigenvalues are both 1.

A = [1 2; 2 1]

and

B = [1 0; 0 1].

The eigenvalues of a matrix and its reduced row echelon form are thus typically not the same.

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FALSE. The eigenvalues of a matrix A are NOT the same as the eigenvalues of the reduced row echelon form of A.

Explanation:

Eigenvalues are scalar values that, when multiplied by an eigenvector, produce the same effect as the original matrix A on that eigenvector. Mathematically, for a matrix A and its eigenvector v, Av = λv, where λ is the eigenvalue.

Row echelon form is a matrix transformation using elementary row operations to obtain a triangular form. The reduced row echelon form (RREF) is a further simplified version of the row echelon form.

However, transforming a matrix A to its RREF changes the properties of the matrix, and therefore the eigenvalues of matrix A and its RREF are generally not the same. The eigenvalues are preserved only under specific matrix transformations, such as similarity transformations, which row echelon form is not one of them.

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The 90% confidence interval for the population mean gpa is [2.545,3.114].

The confidence interval for population mean is:

µ ± z* . σ/√n

Where, μ is population mean, σ is standard deviation, z* is the value of z-score and n is number of samples.

The z-score value for 90% confidence interval is 1.645.

The average gpa is found to be 2.83, so μ=2.83. The variance is 0.45, it means

σ² = 0.45

σ = √0.45

σ = 0.67

The 90% confidence interval for population mean is:

2.83 ± 1.645 . 0.67/√15

[2.83 - 1.645.0.67/√15 , 2.83+1.645.0.67/√15]

[2.545 , 3.114]

Therefore, the 90% confidence interval for the population mean gpa is :

[2.545 , 3.114]

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Answer:

37

Step-by-step explanation:

2 more than a multiple of 7:

multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91

multiples of 7 plus 2: 9, 16, 23, 30, 37, 44, 58, 65, 72, 79, 86, 93

digits add to 10: 37 (3 + 7 = 10)

Answer:

31.55

Step-by-step explanation:

First the original value minus the decreased value

then the new anwser over the original value times by 100

31.55%

$382 .............. 100%

$261.50 ..............x%

x = 261.50×100/382

= 26150/382

= 68.45%

To create a sign chart for the polynomial function F(x) = -(x+5)(x-2)(2x-4)(x-4)², we will examine the intervals defined by the critical points and the zeros of the function.

1. Determine the critical points  -

  - The critical points occur where the factors of the polynomial change sign.

  - The critical points are x = -5, x = 2, x = 4, and x = 4 (repeated).

2. Select test points within each interval  -

  - To evaluate the sign of the polynomial at each interval, we choose test points.

  - Common choices for test points include values less than the smallest critical point, between critical points, and greater than the largest critical point.

  - Let's choose test points  -  x = -6, x = 0, x = 3, and x = 5.

3. Evaluate the sign of the polynomial at each test point

  - Plug in the test points into the polynomial and determine the sign of the expression.

The sign chart for F(x) = -(x+5)(x-2)(2x-4)(x-4)² would look like this

Intervals              Test Point         Sign

-∞ to -5                  -6                    -

-5 to 2                   0                       +

2 to 4                    3                      -

4 to ∞                    5                      +

Note  - The signs in the "Sign" column indicate whether the polynomial is positive (+) or negative (-) in each interval. See the attached sign chart.

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Answer:

insertlinkhere or as in translates 1n53rtl1nkh3r3

Step-by-step explanation:

Answer:

See below ~

Step-by-step explanation:

Subpart (e) :

⇒ New volume = 2 x 3 x 2 x 4 x 2 x 5

⇒ New volume = 8 x 60

⇒ New volume = 480 in³

=============================================================

Subpart (f) :

⇒ New volume : Original volume

⇒ 480 : 60

⇒ 1 : 8

=============================================================

Subpart (g) :

Step 1

⇒ SA = 2(6 × 9 + 9 × 11 + 6 × 11)

⇒ SA = 2(54 + 99 + 66)

⇒ SA = 2(219)

⇒ SA = 438 in²

Step 2

⇒ V = 6 × 9 × 11

⇒ V = 54 × 11

⇒ V = 594 in³

Step 3

⇒ New SA = 2 x 438

⇒ New SA = 876 in²

Step 4

⇒ New : Original

⇒ 876 : 438

⇒ 2 : 1

Step 5

⇒ New V = 2³ x 594

⇒ New V = 8 × 594

⇒ New V = 4752 in³

Step 6

⇒ New : Original

⇒ 4752 : 594

⇒ 8 : 1

On solving the provided question, by properties of parallel lines we got to know that -  1 - A  2 - C  3 - B  4 - D

When two or more lines cross the intersection line, alternate exterior angles result. These angles are created on a number of sides outside the transverse lines. Any two parallel lines that cross one another create two angles with the horizontal line. In the area between the parallel lines, interior angles are formed, while alternate exterior angles are formed in the area outside the parallel lines.

Matches are -

1 - A

2 - C

3 - B

4 - D

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Answer:

2/3 + 1/6 + 2/8

=(2×8+1×4+2×3)/24

=(16+4+6)/24

=26/24

=13/12

*mark me brainliest

Given the number of revolutions, final velocity, and initial velocity, we can determine the angular acceleration of the tires in rad/s2 using the following formula:

\($$\Delta v = v_f - v_i$$$$\alpha = \frac{\Delta v}{t} = \frac{2\pi \Delta N}{t}$$$$t = \frac{\Delta N}{f}$$$$\Delta v = \frac{\Delta S}{t}$$$$\Delta S = \frac{\pi d\Delta N}{1rev}$$$$\alpha = \frac{\pi d\Delta N}{2t} = \frac{\pi d\Delta N f}{2\Delta N}$$Where: $\alpha$ = angular acceleration $\Delta v$ = change in velocity $v_f$ = final velocity $v_i$ = initial velocity $\Delta S$ = change in distance $d$ = diameter $t$ = time $\Delta N$ = a number of revolutions $f$ =\)frequencyWe are given: Diameter of tire = 0.80 mNumber of revolutions = 69 revolutionsFinal speed, v_f = 45 kphInitial speed, v_i = 95 kphWe are to determine the angular acceleration of the tires in rad/s2. We have:Diameter, d = 0.80 number of revolutions, ΔN = 69 revolutionsFinal speed, v_f = 45 kphInitial speed,\(v_i = 95 kph$$\Delta v = v_f - v_i = 45 - 95 = -50\)kph$$Taking the absolute value of the difference, we have:\($$\Delta v = 50 kph = \frac{50}{3.6} m/s = 13.89 m/s$$$$f = \frac{v_f}{\Delta N} = \frac{45}{69} rev/s$$$$t = \frac{\Delta N}{f} = \frac{69}{45/69} s = 106.2 s$$$$\alpha = \frac{\pi d\Delta N f}{2\Delta N} = \frac{\pi d f}{2}$$$$\alpha = \frac{\pi (0.8)(45/69)}{2} rad/s^2 \approx 1.1 rad/s^2$$\)

Therefore, the angular acceleration of the tires in rad/s2 is approximately 1.1 rad/s2.

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As pert the t-test there is sufficient evidence that the average family size in this school district differ from the national average.

Let us assume null and alternate hypothesis.

H₀ = Average family size is 3.18

Hₐ = Average family size is not 3.18

The value of t = (Sample Mean - 3.18) / s/√n

where,

Sample Mean = 3.83

n = 24

s = 1.43

So, t = (3.83 - 3.18) / (1.43 / √24)

t = 2.2268

Degree of freedom = n - 1 = 23

The p-value = 0.0360

As the p-value <0.05, so the null hypothesis will be rejected.

Therefore, there is enough proof to conclude that this school district's average family size is different from the national average.

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Hello there. To solve this question, we'll find how many times George's grandpa is older than his grandson.

First, this is also called the ratio of the ages between George and his grandpa.

To find this, we divide their ages in the order Grandpa : Grandson as following:

Simplify the fraction by a factor of 8

This is how many times George's grandpa is older than him.

To find the derivative of the function h(z) = (ba*z^2)^2, where a and b are constants, we can use the chain rule.

Let's apply the chain rule step by step:

Step 1: Take the derivative of the outer function with respect to its argument (z^2).

d/dz [ (ba*z^2)^2 ] = 2 * (ba*z^2) * d/dz [ ba*z^2 ]

Step 2: Take the derivative of the inner function with respect to its argument (z^2).

d/dz [ ba*z^2 ] = 2 * ba * z

Step 3: Substitute the derivative of the inner function back into the expression from Step 1.

2 * (ba*z^2) * (2 * ba * z) = 4 * (ba)^2 * z^3

Therefore, the derivative of h(z) = (ba*z^2)^2 with respect to z is 4 * (ba)^2 * z^3.

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Answer:

Inequality: y <= 15x

Step-by-step explanation:

y <= 15x

This means the profit per items is no more than 15 per item

$ 58.50

Step-by-step explanation:

because 39 Divided by 6 is 6.5 and 6.5 × 9 it's 58 50