When would you need to arrange polynomials

Answer:

One solution

Step-by-step explanation:

99% of the time, linear equations (equations that have the first degree) have only one solution. However, it's always good to check.

6 - 3x = 12 - 6x

6 = 12 - 3x

-3x = -6

x = 2

As you can see, only one solution. Hope this helps!

Answer:

Step-by-step explanation:

A graphing calculator can perform logarithmic regression, as can a spreadsheet. The least-squares best fit log curve is about ...

  y ≈ 33.7·ln(x) -45.9

The value of x estimated to make y = 5.2 is about 4.6.

Step-by-step explanation:

to be honest I'm not sure how to do

Answer:

a) Mean = 0.125 inch

Standard deviation = 0.13975 inch

b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25) = 0.18673

c) Probability that the door does not fit in the casing = P(X < 0) = 0.18673

Step-by-step explanation:

Let the distribution of the width of the casing be X₁ (μ₁, σ₁²)

Let the distribution of the width of the door be X₂ (μ₂, σ₂²)

The distribution of the difference between the width of the casing and the width of the door = X = X₁ - X₂

when two independent normal distributions are combined in any manner, the resulting distribution is also a normal distribution with

Mean = Σλᵢμᵢ

λᵢ = coefficient of each disteibution in the manner that they are combined

μᵢ = Mean of each distribution

Combined variance = σ² = Σλᵢ²σᵢ²

λ₁ = 1, λ₂ = -1

μ₁ = 24 inches

μ₂ = 23 7/8 inches = 23.875 inches

σ₁² = (1/8)² = (1/64) = 0.015625

σ₂ ² = (1/16)² = (1/256) = 0.00390625

Combined mean = μ = 24 - 23.875 = 0.125 inch

Combined variance = σ² = (1² × 0.015625) + [(-1)² × 0.00390625] = 0.01953125

Standard deviation = √(Variance) = √(0.01953125) = 0.1397542486 = 0.13975 inch

b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25)

This is a normal distribution problem

Mean = μ = 0.125 inch

Standard deviation = σ = 0.13975 inch

We first normalize/standardize 0.25 inch

The standardized score of any value is that value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (0.25 - 0.125)/0.13975 = 0.89

P(X > 0.25) = P(z > 0.89)

Checking the tables

P(x > 0.25) = P(z > 0.89) = 1 - P(z ≤ 0.89) = 1 - 0.81327 = 0.18673

c) Probability that the door does not fit in the casing

If X₂ > X₁, X < 0

P(X < 0)

We first normalize/standardize 0 inch

z = (x - μ)/σ = (0 - 0.125)/0.13975 = -0.89

P(X < 0) = P(z < -0.89)

Checking the tables

P(X < 0) = P(z < -0.89) = 0.18673

Hope this Helps!!!

Answer:

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

Step-by-step explanation:

If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -

x = ( 30 cos 20° )( time ),

y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2

To determine " ( 30 cos 20° )( time ) " you would do the following calculations -

( x = 30 * 0.93... = ( About ) 28.01t

This represents our horizontal distance, respectively the vertical distance should be the following -

y = 30 * 0.34 - 4.9t^2,

( y = ( About ) 10.26t - 4.9t^2 + 2

In other words, our solution should be,

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

These are are parametric equations

Answer: D

Step-by-step explanation:

According to the question, Silver Lake has a population of 114,000. The population is decreasing at a rate of 1.5% each year

The initial population Po = 114000

Rate = 1.5% = 0.015

The declining population formula will be:

P = Po( 1 - R%)x^2

The decay formula

Since the population is decreasing, take away 0.015 from 1

1 - 0.015 = 0.985

Substitutes all the parameters into the formula

P(s) = 114000(0.985)x^2

P(s) = 114000× 0985x^2

The correct answer is written above.

Since option A does not have square of x, we can therefore conclude that the answer is D - non of the choices is correct.

Answer:

Hey there!

There are a few ways you could solve this problem, but the easiest would to be writing an equation.

You could say-

2.3x=69

Divide by 2.3

x=30

Hope this helps :)

Answer:

30

Step-by-step explanation:

the answer is 30 bc increasing something by 130% is multiplying it by 2.3 so technically you have to divide 69 by 2.3 which equals to 30

Answer:

16 ounces = 1 pound

Step-by-step explanation:

You would just do 18/16 = 1.125 pounds. There are always 16 ounces in a pound, so it always works like this

Answer:

80202

Step-by-step explanation:

Simply add according to number value:

200 - 2 goes into hundreds place

2 - 2 goes into ones place

80000 - 8 goes into ten-thousands place

Answer:

  [tex]\dfrac{4x^2+12x+5}{6x^2-3x}[/tex]

Step-by-step explanation:

Invert the denominator and multiply.

  [tex]\dfrac{2x+5}{3x}\div\dfrac{2x-1}{2x+1}=\dfrac{2x+5}{3x}\cdot\dfrac{2x+1}{2x-1}\\\\=\dfrac{(2x+5)(2x+1)}{(3x)(2x-1)}=\boxed{\dfrac{4x^2+12x+5}{6x^2-3x}}\qquad\text{matches choice A}[/tex]

Answer:

[tex]\frac{4x^2+12x+5}{6x^2-3x}[/tex]

Step-by-step explanation:

After using the reciprocal of the second term, the denominator will multiply out to be [tex]6x^2-3x[/tex]. There is only one option with that as the denominator so it must be the correct answer.

Answer:

The frequency table is shown below.

Step-by-step explanation:

The data set arranged ascending order is:

S = {33 , 34 , 39 , 48 , 49 , 50 , 53 , 54 , 55 , 56 , 58 , 58,  60 , 63 , 64 , 65 , 70 , 71 , 74 , 84}

It is asked to use the minimum value from the data set as the lower class limit for the first row.

So, the lower class limit for the first class interval is 33.

To determine the class width compute the range as follows:

[tex]\text{Range}=\text{Maximum}-\text{Minimum}[/tex]

          [tex]=84-33\\=51[/tex]

The number of classes requires is 5.

The class width is:

[tex]\text{Class width}=\frac{Range}{5}=\frac{51}{2}=10.2\approx 10[/tex]

So, the class width is 10.

The classes are:

33 - 42

43 - 52

53 - 62

63 - 72

73 - 82

83 - 92

Compute the frequencies of each class as follows:

Class Interval                  Values                        Frequency

   33 - 42                      33 , 34 , 39                             3

   43 - 52                      48 , 49 , 50                            3

   53 - 62          53 , 54 , 55 , 56 , 58 , 58,  60              7

   63 - 72                 63 , 64 , 65 , 70 , 71                      5

   73 - 82                              74                                  1

   83 - 92                             84                                   1

   TOTAL                                                                   20

Compute the relative frequencies as follows:

Class Interval          Frequency        Relative Frequency

   33 - 42                        3                   [tex]\frac{3}{20}\times 100\%=15\%[/tex]

   43 - 52                        3                   [tex]\frac{3}{20}\times 100\%=15\%[/tex]

   53 - 62                        7                   [tex]\frac{7}{20}\times 100\%=35\%[/tex]

   63 - 72                        5                   [tex]\frac{5}{20}\times 100\%=25\%[/tex]

   73 - 82                         1                   [tex]\frac{1}{20}\times 100\%=5\%[/tex]

   83 - 92                         1                   [tex]\frac{1}{20}\times 100\%=5\%[/tex]

   TOTAL                        20                          100%

Answer:

(a) The value of a is 53.35.

(b) The value of a is 38.17.

(c) The value of a is 26.95.

(d) The value of a is 25.63.

(e) The value of a is 12.06.

Step-by-step explanation:

The probability density function of X is:

[tex]f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}[/tex]

Here, 22 < X < 55.

(a)

Compute the value of a as follows:

[tex]P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35[/tex]

Thus, the value of a is 53.35.

(b)

Compute the value of a as follows:

[tex]P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17[/tex]

Thus, the value of a is 38.17.

(c)

Compute the value of a as follows:

[tex]P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95[/tex]

Thus, the value of a is 26.95.

(d)

Compute the value of a as follows:

[tex]P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63[/tex]

Thus, the value of a is 25.63.

(e)

Compute the value of a as follows:

[tex]P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06[/tex]

Thus, the value of a is 12.06.

Answer:

There are 65 dimes. There are 55 nickels.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

x is the number of dimes

y is the number of nickels.

There are a total of 120 coins in the bag

This means that x + y = 120.

The total value of the coins is $9.25.

The dime is worth $0.10 and the nickel is worth $0.05. So

0.1x + 0.05y = 9.25

System:

[tex]x + y = 120[/tex]

[tex]0.1x + 0.05y = 9.25[/tex]

From the first equation:

[tex]y = 120 - x[/tex]

Replacing in the second:

[tex]0.1x + 0.05y = 9.25[/tex]

[tex]0.1x + 0.05(120 - x) = 9.25[/tex]

[tex]0.1x + 6 - 0.05x = 9.25[/tex]

[tex]0.05x = 3.25[/tex]

[tex]x = \frac{3.25}{0.05}[/tex]

[tex]x = 65[/tex]

[tex]y = 120 - x = 120 - 65 = 55[/tex]

There are 65 dimes. There are 55 nickels.

Answer:

Question 2

Step-by-step explanation:

2) The time when she woke up was -  3° C

During nature walk, temperature got 3° C warmer than when she woke up.

So, temperature during nature walk = - 3 + 3 = 0° C

Answer:

The answer is "Option A"

Step-by-step explanation:

The valid linear programming language equation can be defined as follows:

Equation:

[tex]\Rightarrow \ Min\ 4x + 3y + (\frac{2}{3})z[/tex]

The description of a linear equation can be defined as follows:

It is an algebraic expression whereby each term contains a single exponent, and a single direction consists in the linear interpolation of the equation.

Formula:

[tex]\to \boxed{y= mx+c}[/tex]

Answer:

5.35 more miles

Answer:

5.35 miles to the finish line

Step-by-step explanation:

Step one

13.1-7.75=

5.35

Answer:

B

Step-by-step explanation:

A is not a function because the same x value is repeated twice with different y values. The same goes for C and D so the answer is C.

Answer:

B.

Step-by-step explanation:

Well a relation is a set of points and a function is a relation where every x value corresponds to only 1 y value.

So lets see which x values in these relations have only 1 y value.

A. Well a isn’t a function because the number 3 which is a x value had two y values which are -1 and 4.

B. This relation is a function because there are no similar x values.

C. This is not a function because the x value 4 has two y values which are 4 and -3.

D. This is not a function because the number 1 has 2 and 3 as y values.

Answer:

When describing a property line drawn down the center of a creek bed

Answer:

81.23 ft and 77.88 ft long

Step-by-step explanation:

The sum of the internal angles of a triangle is 180 degrees, the missing angle is:

[tex]a+b+c=180\\a+23+22=180\\a=135^o[/tex]

According to the Law of Sines:

[tex]\frac{A}{sin(a)}= \frac{B}{sin(b)}= \frac{C}{sin(c)}[/tex]

Let A be the side that is 147 feet long, the length of the other two sides are:

[tex]\frac{A}{sin(a)}= \frac{B}{sin(b)}\\B=\frac{sin(23)*147}{sin(135)}\\B=81.23\ ft\\\\\frac{A}{sin(a)}= \frac{C}{sin(c)}\\C=\frac{sin(22)*147}{sin(135)}\\C=77.88\ ft[/tex]

The other two sides are 81.23 ft and 77.88 ft long

Answer:

B. g(x) = (x-2)^2 +1

Step-by-step explanation:

When you see this type of equation your get the variables H and K in a quadratic equation. In this case the (x-2)^2 +1  is your H. The (x-2)^2 +1 is your K.

For the H you always do the opposite so in this case instead of going to the left 2 times you go to the right 2 times (affects your x)

For the K you go up or down which in this case you go up one (affects your y)

And that's how you got your (2,1) as the center of the parabola

-Hope this helps :)

Answer:

(a)0.5

(b)0.17

(c)0.625

(b)0.375

Step-by-step explanation:

Pr(E)=0.6​

Pr(F)=0.2

[tex]Pr(E\cap F)=0.1.[/tex]

(a)Pr (E|F )

[tex]Pr (E|F )=\dfrac{Pr(E \cap F)}{Pr(F)} \\=\dfrac{0.1}{0.2}\\\\=0.5[/tex]

(b)Pr (F|E )

[tex]Pr (F|E )=\dfrac{Pr(E \cap F)}{Pr(E)} \\=\dfrac{0.1}{0.6}\\\\=0.17[/tex]

(c)Pr (E|F')​

Pr(F')=1-P(F)

=1-0.2=0.8

[tex]Pr(E \cap F')=P(E)-P(E\cap F)\\=0.6-0.1\\=0.5[/tex]

Therefore:

[tex]Pr (E|F' )=\dfrac{Pr(E \cap F')}{Pr(F')} \\=\dfrac{0.5}{0.8}\\\\=0.625[/tex]

(d)Pr(E'|F')​

[tex]P(E'\cap F')=P(E \cup F)'\\=1-P(E \cup F)\\=1-[P(E)+P(F)-P(E\cap F)]\\=1-[0.6+0.2-0.1]\\=1-0.7\\=0.3[/tex]

Therefore:

[tex]Pr (E'|F' )=\dfrac{Pr(E' \cap F')}{Pr(F')} \\=\dfrac{0.3}{0.8}\\\\=0.375[/tex]

Answer:  [tex]H_0:\mu=421[/tex]

[tex]H_a : \mu\neq421[/tex]

Step-by-step explanation:

Given: A 37 bag sample had a mean of 421 grams.

Let [tex]\mu[/tex] be the population mean.

Then, the null hypothesis would be:

[tex]H_0:\mu=421[/tex]

whereas the alternative hypothesis would be:

[tex]H_a : \mu\neq421[/tex]

Answer:

Step-by-step explanation:

Let x represent the length of the side not parallel to the partition. Then the length of the side parallel to the partition is ...

  y = (300 -2x)/3

And the enclosed area is ...

  A = xy = x(300 -2x)/3 = (2/3)(x)(150 -x)

This is the equation of a parabola with x-intercepts at x=0 and x=150. The line of symmetry, hence the vertex, is located halfway between these values, at x=75.

The maximum area is enclosed when the dimensions are ...

  50 ft by 75 ft

That maximum area is 3750 square feet.

_____

Comment on the solution

The generic solution to problems of this sort is that half the fence (cost) is used in each of the orthogonal directions. Here, half the fence is 150 ft, so the long side measures 150'/2 = 75', and the short side measures 150'/3 = 50'. This remains true regardless of the number of partitions, and regardless if part or all of one side is missing (e.g. bounded by a barn or river).

Answer:

We conclude that the rate of inaccurate orders is greater than ​10%.

Step-by-step explanation:

We are given that in a study of the accuracy of fast food​ drive-through orders, one restaurant had 40 orders that were not accurate among 307 orders observed.

Let p = population proportion rate of inaccurate orders

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 10%     {means that the rate of inaccurate orders is less than or equal to ​10%}

Alternate Hypothesis, [tex]H_A[/tex] : p > 10%      {means that the rate of inaccurate orders is greater than ​10%}

The test statistics that will be used here is One-sample z-test for proportions;

                          T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of inaccurate orders = [tex]\frac{40}{307}[/tex] = 0.13

           n = sample of orders = 307

So, the test statistics =  [tex]\frac{0.13-0.10}{\sqrt{\frac{0.10(1-0.10)}{307} } }[/tex]  

                                    =  1.75

The value of z-test statistics is 1.75.

Also, the P-value of the test statistics is given by;

            P-value = P(Z > 1.75) = 1 - P(Z [tex]\leq[/tex] 1.75)

                          = 1 - 0.95994 = 0.04006

Now, at 0.05 level of significance, the z table gives a critical value of 1.645  for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 1.75 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the rate of inaccurate orders is greater than ​10%.

Answer:

(A)Step 1; it should be m∠m + m∠n + m∠o = 180 degrees (sum of angles in a triangle)

Step-by-step explanation:

The steps followed by the student are:

Step 1: m∠m + m∠n + m∠p = 180 degrees (sum of angles of a triangle)

Step 2: m∠p + m∠o = 180 degrees (adjacent supplementary angles)

Step 3: Therefore, m∠m + m∠n + m∠o = m∠o + m∠p

Step 4: So, m∠m + m∠n = m∠p

We observe that the student made a mistake in Step 1, it should be m∠m + m∠n + m∠o = 180 degrees (sum of angles in a triangle).

p is outside the triangle, therefore it cannot form one of the angles in the triangle.

Answer:

We can use the cross products property.

11/8 = y / 13

8y = 11 * 13

y = 11 * 13 / 8 = 17.875

Answer:

y=17.875

Step-by-step explanation:

[tex]\frac{11}{8} = \frac{y}{13}[/tex]

11(13)=8y

143=8y

y=17.875

Answer:

x = 0,00375 mm

Step-by-step explanation:

a) El factor de ampliación es 400/1   es decir el tamaño real se verá ampliado 400 veces mediante el uso del microscopio

b) De acuerdo a lo establecido en la respuesta a la pregunta referida en a (anterior) podemos establecer una regla de tres, según:

Si al microscopio el tamaño de la célula es 1,5 mm, cual será el tamaño verdadero ( que es reducido 400 en relación al que veo en el microscopio)

Es decir     1,5 mm      ⇒    400

                    x (mm)    ⇒       1 (tamaño real de la célula)

Entonces

x  =  1,5 /400

x = 0,00375 mm

Answer:

0

Step-by-step explanation:

multiply the negative thru the right part of the equation so, 5j+5-5j-5. The 5j and the 5 than cancel out with each other. Hope this helps!

Answer:

0

Explanation:

step 1 - remove the parenthesis from the expression

(5j + 5) - (5j + 5)

5j + 5 - 5j - 5

step 2 - combine like terms

5j + 5 - 5j - 5

5j - 5j + 5 - 5

0 + 0

0

therefore, the simplified form of the given expression is 0.

Answer:

The equation for this ellipse is [tex]\frac{x^{2}}{64} + \frac{y^{2}}{16} = 1[/tex].

Step-by-step explanation:

The standard equation of the ellipse is described by the following expression:

[tex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1[/tex]

Where [tex]a[/tex] and [tex]b[/tex] are the horizontal and vertical semi-axes, respectively. Given that major semi-axis is horizontal, [tex]a > b[/tex]. Then, the equation for this ellipse is written in this way: (a = 8, b = 4)

[tex]\frac{x^{2}}{64} + \frac{y^{2}}{16} = 1[/tex]

The equation for this ellipse is [tex]\frac{x^{2}}{64} + \frac{y^{2}}{16} = 1[/tex].