How to simplify this expression??

How To Simplify This Expression??

Answers

Answer 1

Answer :

[tex] \frac{2 {x}^{3} + 7 {x}^{2} + 3x - 4}{ {x}^{3} + 3 {x}^{2} + x - 1} [/tex]

Step-by-step-explanation :

I did the explanation in the picture.

How To Simplify This Expression??

Related Questions

5x+4(-x-2)=-5x+2(x-1)+12

Answers

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Three out of every ten dentists recommend a certain brand of fluoride toothpaste. Which assignment of random digits would be used to simulate the random sampling of dentists who prefer this fluoride toothpaste?

Answers

Answer:

eddfdgdccggģdffcdrrfxddxcvgfx

Transform the given parametric equations into rectangular form. Then identify the conic.

Answers

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

 

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente

Answers

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

X = 1mL


Write in words how we would say the following
3 square​

Answers

Answer:

Three to the second power

Step-by-step explanation:

Hey there!

3 square

Can be written as the following,

Three to the second power

Hope this helps :)

find the perimeter of a square of sides 10.5cm​

Answers

Answer:

Perimeter = 42 cm

Step-by-step explanation:

A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.

Answer:

42 cm

Step-by-step explanation:

Side of square = 10.5 cm (given)

Perimeter of square = Side X 4

                                  = 10.5 X 4

                                  = 42 cm

HOPE THIS HELPED YOU !

:)

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal places.)

a. x=5
b. x <= 5
c. x>=6

Answers

Answer:

[tex]\mathbf{P(X=5) =0.0888}[/tex]    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

[tex]\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]

[tex]\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]

where;

n = 8 and π = 0.36

For x = 5

The probability [tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}[/tex]

[tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}[/tex]

[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}[/tex]

[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}[/tex]

[tex]\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}[/tex]

[tex]\mathtt{P(X=5) =0.0887645}[/tex]

[tex]\mathbf{P(X=5) =0.0888}[/tex]     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)[tex]\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})[/tex]

[tex]{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +[/tex][tex]\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )[/tex]

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

A machine used to fill​ gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be​ reset? Explain your reasoning. ▼ Yes No ​, it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ​ounces, because it ▼ lies does not lie within the range of a usual​ event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.

Answers

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Use Lagrange multipliers to minimize the function subject to the following two constraints. Assume that x, y, and z are nonnegative. Question 18 options: a) 192 b) 384 c) 576 d) 128 e) 64

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option C is the correct option

Step-by-step explanation:

From the question we are told that

   The equation is  [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]

    The constraint is  [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]

Now using Lagrange multipliers  we have that  

   [tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]

=>       [tex]x = \frac{ \lambda }{2}[/tex]

          [tex]y = \frac{ \lambda }{2}[/tex]

         [tex]z = \frac{ \lambda }{2}[/tex]

From the constraint  we have

      [tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]

=>   [tex]\frac{3 \lambda }{2} = 24[/tex]

=>   [tex]\lambda = 16[/tex]

substituting for x, y, z

=>   x =  8

=>  y =  8

=>   z =  8        

Hence

    [tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]

    [tex]f (8, 8 , 8 ) = 192[/tex]

 

which of the following not between -10 and -8

-17/2
-7
-9
-8.5​

Answers

The answer is -7 because -17/2=-8.5 and 9 and 8.5 are both in between -10 and -8

Answer:

-7

Step-by-step explanation:

This is best read on the number line.

Look at the picture.

[tex]-\dfrac{17}{2}=-8\dfrac{1}{2}=-8.5[/tex]

A regular polygon inscribed in a circle can be used to derive the formula for the area of a circle. The polygon area can be expressed in terms of the area of a triangle. Let s be the side length of the polygon, let r be the hypotenuse of the right triangle, let h be the height of the triangle, and let n be the number of sides of the regular polygon. polygon area = n(12sh) Which statement is true? As h increases, s approaches r so that rh approaches r². As r increases, h approaches r so that rh approaches r². As s increases, h approaches r so that rh approaches r². As n increases, h approaches r so that rh approaches r².

Answers

Answer:

Option (D)

Step-by-step explanation:

Formula to get the area of a regular polygon in a circle will be,

Area = [tex]n[\frac{1}{2}\times (\text{Base})\times (\text{Height})][/tex]

        = [tex]n[\frac{1}{2}\times (\text{s})\times (\text{h})][/tex]

Here 'n' is the number of sides.

If n increases, h approaches r so that 'rh' approaches r².

In other words, if the number of sides of the polygon gets increased, area of the polygon approaches the area of the circle.

Therefore, Option (4) will be the answer.

In this exercise it is necessary to have knowledge about polygons, so we have to:

Letter D

Then using the formula for the area of ​​a regular polygon we find that:

[tex]A=n(1/2*B*H)\\=n(1/2*S*H)[/tex]

So from this way we were not able to identify the option that best corresponds to this alternative.

See more about polygons at  brainly.com/question/17756657

cooks are needed to prepare for a large party. Each cook can bake either 5 Large cakes or 14 small cakes per hour . The kitchen is available for 3 hours and 29 large cakes and 260 cakes need to be baked . How many cooks are required to bake the required number of cakes during the time the kitchen is available?​

Answers

it was all about equating some values

to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.

Let's determine the number of cooks required to bake the required number of cakes during the available time.

We have the following information:

- Each cook can bake either 5 large cakes or 14 small cakes per hour.

- The kitchen is available for 3 hours.

- We need to bake 29 large cakes and 260 cakes in total.

First, let's calculate the number of large cakes that can be baked by one cook in 3 hours:

1 cook can bake 5 large cakes/hour × 3 hours = 15 large cakes.

Next, let's calculate the number of small cakes that can be baked by one cook in 3 hours:

1 cook can bake 14 small cakes/hour × 3 hours = 42 small cakes.

Now, let's calculate the number of large cakes that can be baked by all the cooks in 3 hours:

Total number of large cakes = Number of cooks × Large cakes per cook per 3 hours

We need to bake 29 large cakes, so:

29 = Number of cooks × 15

Number of cooks = 29 / 15 ≈ 1.93

Since we can't have a fraction of a cook, we need to round up to the nearest whole number. Therefore, we need at least 2 cooks to bake the required number of large cakes.

Similarly, let's calculate the number of small cakes that can be baked by all the cooks in 3 hours:

Total number of small cakes = Number of cooks × Small cakes per cook per 3 hours

We need to bake 260 small cakes, so:

260 = Number of cooks × 42

Number of cooks = 260 / 42 ≈ 6.19

Again, rounding up to the nearest whole number, we need at least 7 cooks to bake the required number of small cakes.

Since we need to satisfy both requirements for large and small cakes, we choose the larger number of cooks required, which is 7 cooks.

Therefore, to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.

Learn more about work here

https://brainly.com/question/13245573

#SPJ2

Determine the value of x in the figure. Question 1 options: A) x = 90 B) x = 85 C) x = 45 D) x = 135

Answers

Answer:

A.) x=90°

Step-by-step explanation:

Note:

The triangle shown is an isosceles triangle, which means that it has 2 congruent sides (as shown by the small intersecting lines), and this also means that it has two congruent angles.

We are given an angle measure adjacent to one of the missing angles. These two form supplementary angles, which means that they're sum is equal to 180°, or a straight line. So, to find:

[tex]180=135+y[/tex]

y is the unknown angle. Solve for y:

[tex]180-135=y\\\\y=45[/tex]

y is 45°. Since this and the other angle are congruent, add:

[tex]45+45=90[/tex]

Note:

Triangles angles will always add up to a total of 180°.

To find the missing angle x°, use:

[tex]180=a+b+c[/tex]

These are the angles in a triangle. Substitute any known values and solve:

[tex]180=45+45+x\\\\180=90+x\\\\180-90=x\\\\x=90[/tex]

The missing angle x° is 90°.

:Done

The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:

Answers

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

   

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

       

 

Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?
A)Right
B)Obtuse
C)Can't be determined
D) Acute

Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?
A)0.33 feet
B)3.75 feet
C)3 feet
D)5 feet

Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?
A)Acute
B)Right
C)Can't be determined
D)Obtuse

Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?
A)21.34 ft.
B)21.93 ft.
C)27.73 ft.
D)19.21 ft.

Answers

Answer:

Question 1 = D) Acute

Question 2 = C)3 feet

Question 3 = D) Obtuse

Question 4 = C)27.73 ft.

Step-by-step explanation:

Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths

5 = a, 6 = b and c = 7

a² + b² = c²

5² + 6² = 7²

25 + 36 = 49

61 = 49

61 ≠ 49, Hence 61 > 49

Therefore, this is an Acute Triangle

Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?

This is question that deals with proportion.

The formula to solve for this:

Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person

Height of the statue = 15 feet

Length of the shadow of the person = 20 feet

Height of the person = unknown

Length of the shadow of the person = 4

15/ 20 = Height of the person/4

Cross Multiply

15 × 4 = 20 × Height of the person

Height of the person = 15 × 4/20

= 60/20

Height of the person = 3 feet

Therefore, the person is 3 feet tall.

Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths 17, 12, 9

9 = a, 12 = b and c = 17

a² + b² = c²

9² + 12² = 17²

81 + 144 = 289

225 = 289

225 ≠ 289

225 < 289

Hence, This is an Obtuse Triangle.

Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?

To calculate how far apart the two friends are we use the formula

Distance = √ ( Length² + Breadth²)

We are given dimensions: 12ft by 25ft

Length = 12ft

Breadth = 25ft

Distance = √(12ft)² + (25ft)²

Distance = √144ft²+ 625ft²

Distance = √769ft²

Distance = 27.730849248ft

Approximately ≈27.73ft

Therefore, the friends are 27.73ft apart.

A senior accounting major at Midsouth State University has job offers from four CPA firms. To explore the offers further, she asked a sample of recent trainees how many months each worked for the firm before receiving a raise in salary. The sample information is submitted to Minitab with the following results:
Analysis of Variance
Source df SS MS F P
Factor 3 28.17 9.39 5.37 0.010
Error 15 26.26 1.75
Total 18 54.43
A) Reject H0 if F >
B) For the 0.05 level of significance, is there a difference in the mean difference in the mean number of months before a raise was anted among the four CPA firms?

Answers

Answer:

A) Reject H0 if F > 5.417

B) we fail to reject the null hypothesis and conclude that we do not have sufficient evidence at 0.05 level of significance to support the claim that there is a difference in the mean number of months before a raise was granted among the four CPA firms

Step-by-step explanation:

A) From the table, we can see that we have df1 = 3 and df2 = 15. And we are given a significance level of α = 0.01

We are also given f-value of 1.75

Thus,from the f-distribution table attached at significance level of α = 0.01 and df1 = 3 and df2 = 15, we have;

F-critical = 5.417

Normally, we reject H0 if F > 5.417

But in this case, F is 1.75 < 5.417 and so we conclude that we do not reject H0 at the 0.01 level of significance

B) for 0.05 level of significance, df1 = 3 and df2 = 15, from the 2nd table attached, we have;

F-critical = 3.2874

Again the f-value is less than this critical one.

Thus, we fail to reject the null hypothesis and conclude that we do not have sufficient evidence at 0.05 level of significance to support the claim that there is a difference in the mean number of months before a raise was granted among the four CPA firms

Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

Answers

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

You are studying for your final exam of the semester up to this point you received 3 exam scores of 61% 62% and 86% to receive a grade of c and the class you must have an average exam score between 70% and 79% for all four exams including the final find the widest range of scores that you can get on the final exam in order to receive a grade of C for the class 63 to 100% 71 to 100% 68 to 97

Answers

There will be a total of 4 test scores including the final exam. To get a 70, the 4 tests need to equal 4 x 70 = 280 points , to be 79, they have to equal 4 x 79 = 316 points.

The 3 already done = 61 + 62 + 86 = 209 points.

The final exam needs to be between :

280 -209 = 71

316 -209 = 107. The answer would be between 71 and 100%



Type the missing number in this sequence:
1,
4,
,64, 256,
1,024

Answers

Answer:

16

Step-by-step explanation:

The sequence is 1, 4,...,64, 256, 1024

Notice that:

● 1 = 2^0

● 4 = 2^2

● 64 = 2^6

● 256 = 2^8

● 1024 = 2^10

Notice that we add 2 each time to the exponent so the missing number is:

● 2^(2+2) = 2^4 = 16

Please help ! I’ll mark you as brainliest if correct.

Answers

Answer:

D = -87Dx = 174Dy = -435Dz = 0(x, y, z) = (-2, 5, 0)

Step-by-step explanation:

The determinant of the coefficient matrix is ...

  [tex]D=\left|\begin{array}{ccc}2&5&3\\4&-1&-4\\-5&-2&6\end{array}\right|\\\\=2(-1)(6)+5(-4)(-5)+3(4)(-2)-2(-4)(-2)-5(4)(6)-3(-1)(-5)\\\\=-12+100-24-16-120-15=\boxed{-87}[/tex]

The other determinants are found in similar fashion after substituting the constants on the right for each of the above matrix columns, in turn.

Those determinants are ...

  [tex]D_x=\left|\begin{array}{ccc}21&5&3\\-13&-1&-4\\0&-2&6\end{array}\right|=174[/tex]

  [tex]D_y=\left|\begin{array}{ccc}2&21&3\\4&-13&-4\\-5&0&6\end{array}\right|=-435[/tex]

  [tex]D_z=\left|\begin{array}{ccc}2&5&21\\4&-1&-13\\-5&-2&0\end{array}\right|=0[/tex]

The solutions are ...

  x = 174/-87 = -2

  y = -435/-87 = 5

  z = 0

That is, (x, y, z) = (-2, 5, 0).

one third multiplied by the sum of a and b

Answers

Answer:

1/3(a+b)

hope it helps :>

a+b/3
This is the answer of ur question

Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale

Answers

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

Answers

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

You are ordering two pizzas. A pizza can be small, medium, large, or extra large, with any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8). How many possibilities are there for your two pizzas

Answers

Answer:

1048576

Step-by-step explanation:

Given the following :

Pizza order :

Size = small, medium, large, or extra large = 4 possible sizes

Toppings = any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8).

Combination of Toppings = 2^8

Four different sizes of pizza = 4

Number of possibilities in ordering for a single pizza :

(4 * 2^8) = 4 * 256 = 1024

Number of possibilities in ordering two pizzas :

(4 * 2^8)^2

(2^2 * 2^8)^2

From indices :

[2^(2+8)]^2

[2^(10)]^2

2^(10*2)

2^20

= 1048576

(21x-3)+21=23x+6 solve​

Answers

Answer:

False

Step-by-step explanation:

You Cnat solve it

Answer:

you cannot solve it

Step-by-step explanation:

false

Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Answers

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Please help me solve for the median !!!

Answers

Answer:

50.93

Step-by-step explanation:

Add up the frequencies:

2 + 5 + 14 + 15 + 21 + 18 + 15 + 9 + 2 = 101

Divide by 2: 101/2 = 50.5

So the median is the 51st number, with 50 below and 50 above.

Add up the frequencies until you find the interval that contains the 51st number.

2 + 5 + 14 + 15 = 36

2 + 5 + 14 + 15 + 21 = 57

So the median is in the group 49.5 − 51.5.  To estimate the median, we use interpolation.  Find the slope of the line from (36, 49.5) to (57, 51.5).

m = (51.5 − 49.5) / (57 − 36)

m = 2/21

So at x = 51:

2/21 = (y − 49.5) / (51 − 36)

y = 50.93

Use the two highlighted points to find the
equation of a trend line in slope-intercept
form.

Answers

Answer: y=(4/3)x+2/3

Step-by-step explanation:

Slope-intercept form is expressed as y=mx+b

First, find the slope (m):

m= rise/run or vertical/horizontal or y/x (found between the highlighted points)

m = 4/3

Second, find b:

Use one of the highlighted points for (x, y)

2=4/3(1)+b

6/3=4/3+b

2/3=b

b=2/3

Plug it into the equation:

You get y=(4/3)x+2/3 :)

10) How many possible outfit combinations come from six shirts, three
slacks, and five ties? *
A 15
B 18
C 30
D 90

Answers

Answer:

The answer is D)90

Hope I helped

A professional soccer player kicked a ball across the field. The ball’s height, in meters, is modeled by the function graphed below. What's the average rate of change between the point when the ball reached its maximum height and the point where it hit the ground?

Answers

Answer:

Hey there!

You can think of the rate of change as the slope of a quadratic function- here we see that it is 9/-3, or - 3.

Let  me know if this helps :)

Answer:

–3 meters per second

Step-by-step explanation:

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