For what values of x is the perimeter of the figure greater than 112 ​meters?

Select the correct choice below and fill in the answer box to complete your choice.

Answer:

113

Step-by-step explanation:

An airline that wishes to measure customer satisfaction chooses ten of its flights at random over the course of a single month and requests that each passenger fill out a survey. A Cluster sample looks like this.

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Answer:

21 m/ph

Step-by-step explanation:

2016 divided by 96

The balloon travels 21 miles in one hour.

In mathematics, division is one kind of operation. The phrases or numbers in this process are split into the same number of parts.

Given, a hot air balloon travels 2016 miles in 96 hours.

And the balloon travels the same number of miles each hour.

To find the same numbers of miles per hour:

We use the division of 2016 by 96.

That means,

2016 / 96

= 21

Therefore, the travel time of the balloon is 21 miles per hour.

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The probability that overbooking occurs is 0.7920 and the probability that the flight has empty seats is 0.0510 respectively.

We remove the 16 regular customers and their seats from consideration. Thus, we consider the 22 - 16 remaining customers = 6 and the 18-16 = 2 seats overbooking means 3 or more of the 6 remaining passengers; each has a 58% chance remaining.  We can either use the binomial distribution formula P(x) = C(6, x) .58^x .42^(6-x) or use packages that provide the result, including cumulative probabilities.

Overbooking = 1 - P(X <= 2)   In Excel, we use =1 - binom.dist(2,6,.58,1)   ; the 1 is for cumulative.

The solution is 0.7920

Probability of empty seats = P(x <= 1)  = binom.dist(1,6,.58,1) = 0.0510

The probability that overbooking occurs is 0.7920 and the probability that the flight has empty seats is 0.0510 respectively.

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Answer:

-5/8

Step-by-step explanation:

5.(-2^-3)

-5. 1/2³

-5. 1/8

-5/8

Answer:

Area = 12a m²

Step-by-step explanation:

Given information,

→ Length = 4a m

→ Width = 3a m

Now we have to,

→ find the area of rectangular park.

Formula we use,

→ Area = L × W

Then the area of rectangle is,

→ L × W

→ 4a × 3a

→ (4 × 3)a

→ 12a m²

Therefore, the area is 12a m².

The amount of sugar, in ounces, that is needed to make 28 ounces of hummingbird nectar is 7 ounces.

Hummingbird nectar is said to be made by using a ratio of 1 part sugar to 3 parts water. For every 1 ounce of sugar therefore, there would be 3 ounces of water.

Put together, this would give 4 ounces. And the percentage of sugar in this 4 ounces would be:

= 1 / 4

= 25 %

If 28 ounces of hummingbird nectar needed to be made, the sugar needed is :

= Percentage of sugar x Ounces of hummingbird nectar

= 25 % x 28

= 7 ounces of sugar

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Finding the values of d and the value of m with the algebraic working will give us the the value of d to be 7 and the value of m to be 3.

Finding the value of d, we know that the sum of the first n terms of an arithmetic series is given by:

S = n/2 * (2a + (n-1)d)

Since the sum of the first n terms is S and the sum of the first 2m terms is S2m, we can set up the following equation:

S = m/2 * (2a + (m-1)d)

S2m = 2m/2 * (2a + (2m-1)d)

Substituting the given values for S and S2m into these equations, we get:

39 = m/2 * (2a + (m-1)d)

320 = 2m/2 * (2a + (2m-1)d)

Solving for d in each equation, we find that d = -7 in the first equation and d = 7 in the second equation. Since d must be a prime number, the only possible value for d is 7.

Now that we know the value of d, we can solve for m. Substituting the value of d back into one of the equations and solving for m, we get:

39 = m/2 * (2a + (m-1)7)

78 = m * (2a + (m-1)7)

78 = m * 2a + 7m^2 - 7m

7m^2 - m - 78 = 0

We can solve for m using the quadratic formula:

m = (-1 +/- sqrt(1^2 - 4*7*(-78)))/(2*7)

= (-1 +/- sqrt(2521))/14

= (-1 + 49)/14 = 3

= (-1 - 49)/14 = -7

Since m must be a positive integer, the only possible value for m is 3.

Therefore, the value of d is 7 and the value of m is 3.

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Answer: 2 times 2 times 2 times 2 times 3 times 3

Step-by-step explanation:

a.) The amount that Sammy will have in his account after 10 years will be= $2,165.8

$2,165.8b.) The value of 'r' required would be = 0.27%

Interest rate is defined as the rate at which an individual receives an amount of money from an investment made over a period of time.

The amount invested by Sam(P) = $1700

The interest rate(R) = 2.74%

The time of investment (T) = 10 years

Simple interest = P×T×R/100

= 1700×10×2.74/100

= 46580/100

= $465.80

Therefore, the amount that Sammy will have in his account after 10 years will be = $1700 + $465.80

= $2,165.8.

For David's account interest rate the following is carried out:

The amount invested by Sam(P) = $1700

The interest rate(R) = r%

The time of investment (T) = 10 years

Simple interest (SI) = $465.80

Using the formula for simple interest;

SI = P×T×R/100

Make R the subject of formula;

R = SI×100/P×T

R = 465.80×100/1700×100

R= 46580/170000

R= 0.27%

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Complete question:

Sam invests $1700 in a saving account that pays a nominal annual rate of interest of 2.74% compounded half-yearly. Sam makes no further payments to, or withdrawals from, this account. (a) Find the amount that Sam will have in his account after 10 years.

David also invests $1700 in a savings account that plays an annual rate of interest of r%, compounded yearly. David makes no further payments or withdrawals from this account. (b) Find the value of 'r' required so that the amount in David's account after 10 years will be equal to the amount in Sam's account.

The answers reasonably approximates the requested value of the sample mean with justification is c. "Bar-w"=13.70 since we can use a normal approximation by the CLT.

Since the sample size is large (n=50), we can use the central limit theorem (CLT) to approximate the sampling distribution of the sample mean with a normal distribution.

The mean of the sampling distribution of the sample mean is equal to the mean of the population, which is 13 feet in this case. The standard deviation of the sampling distribution of the sample mean is equal to the standard deviation of the population divided by the square root of the sample size. This is known as the standard error of the mean, and is denoted by SE(Bar-W).

In this case, SE(Bar-W) = 2.4/sqrt(50) = 0.48.

To find the 98th percentile of a normal distribution with mean 13 and standard deviation 0.48, we can use a standard normal table or a calculator to find that the 98th percentile is approximately 2.05.

Therefore, the value that 98% of samples will have the realized value of Bar-W less than is approximately 13 + 2.05 * 0.48 = 13.70. This means that the correct answer is (c) "Bar-w"=13.70 since we can use a normal approximation by the CLT.

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Answer:

  x = 16 1/3°

Step-by-step explanation:

You have parallel lines with consecutive interior angles marked 7x-28° and 5x+12°, and you want to find the value of x.

At parallel lines, consecutive interior angles, or same-side interior angles, are supplementary. This means the total of the two marked angles is 180°.

  7x -28° +5x +12° = 180°

  12x -16° = 180° . . . . . . . . . . simplify

  12x = 196° . . . . . . . . . add 16°

  x = (49/3)° = 16 1/3° . . . . . . . divide by 12

__

Additional comment

∠3 = 86 1/3°

∠5 = 93 2/3°

What is the greatest common factor of 20 and 40?

20

Answer:

4xy ×3xy -2xy

12xy-2xy

10xy

Answer:

lies between 10 and 11.

Step-by-step explanation:

We are asked to find the values between which  lies.

We know that square root of 100 is 10 and square root of 121 is 11.

Since  is between  and , therefore,  lies between 10 and 11.

The distribution of Z = X/Y is  λ₁ / (λ₁+λ₂).

The probability distribution of random variables is described using the cumulative distribution function. The probability for a discrete, continuous, or mixed variable may be described using it. The cumulative probability for a random variable is calculated by adding the probability density function.

Given:

X and Y are independent exponential random variables with respective parameters λ₁ and λ₂.

To find the distribution of Z = X/Y:

First, we have

[tex]f_x_y[/tex](x,y) = λ₁λ₂[tex]e^{-\lambda_1x[/tex][tex]e^{-\lambda_2y[/tex]

First, we find the cumulative distribution function (CDF) for Z = X/Y.

Derivative of Z, f(z) and put  a = x/y

[tex]F_z[/tex](a)  = P (X/Y ≤ a)

        = P (X ≤ aY)

        = [tex]\int\limits^{\infty}_ {x} \ \int\limits^{ay}_0 {\lambda_1\lambda_2e^{-\lambda_1x}e^{-\lambda_2y}\ dx dy[/tex]

       [tex]= \int\limits^{\infty}_ {0} {\lambda_1\lambda_2e^{-\lambda_2y}\ dy [ -1/{\lambda_1}{e^{-\lambda_1x}]\limits^{ay}_ {0}[/tex]

       [tex]= \lambda_2\int\limits^{\infty}_ {0} {e^{-\lambda_2y} - e^{-y(\lambda_2 + \lambda_1a)} \ dy[/tex]

        [tex]= \lambda_2[ [{1/{\lambda_2}+{\lambda_1a]-{e^{-\lambda_2y} + e^{-y(\lambda_2 + \lambda_1a)}]\limits^{\infty}_ {0}[/tex]

        [tex]= -[({\lambda_2}/{\lambda_2 + \lambda_1) - 1][/tex]

        [tex]= -[({\lambda_2}/{\lambda_2 + \lambda_1a) - 1][/tex]

       [tex]= [({\lambda_1a}/{\lambda_2 + \lambda_1a)][/tex]

So, P(X<Y) = P (X/Y <1)

                 = λ₁ / (λ₁+λ₂)

Therefore, distribution is λ₁ / (λ₁+λ₂).

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Answer: -32

Step-by-step explanation:

To solve the equation -128 = 4x for x, we need to isolate the x variable on one side of the equation. We can do this by dividing both sides of the equation by 4. This gives us:

-128/4 = 4x/4

Dividing -128 by 4 gives us -32, so we have:

-32 = 4x/4

We can simplify this by dividing both sides by 4/4 to get:

-32 = x

Thus, the solution to the equation -128 = 4x is x = -32.

2. It is false that the differential equation is non-linear.

The statement regarding the differential equation y''y + (y')² = sin(x) is false.

3. Statement 3: fourth order and nonlinear differential equation.

For a linear differential equation, the coefficients of y or the derivatives of y are either:

Hence the differential equation for question 2 is linear, as the coefficients of the derivatives are x and -1, neither of which is y, hence the statement is false.

For the differential equation y''y + (y')² = sin(x), we have that:

For the differential equation in item 3, we have that:

Hence statement 3 is correct.

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A sample of 24 pieces of work is considered for the checking the quality of the work,

a) Upper control limit of 99.73% confidence interval is 4

Lower control limit of 99.73% confidence interval is Zero.

The process is in control because all the values of the blemishes per unit are lie within the control limits .

The Control limits consist of two limits lower control limit and upper control limit within which the value of the statistical observations of a product are expected to lie. If the values lie outside these limits, then the process is called out of statistical control. We have,

Kate Drew has been hand-painting wooden Christmas ornaments for several years.

Sample size = 24 pieces

confidence level = 99.73% = 0.9973

The number of blemishes per unit is given as:

1,1,2,3,1,0,2,0,0,1,3,1,0,1,2,0,0,2,1,2,3,2,1,1.

The control limits are to be computed for the number of defects per piece . Let us consider the number of defects per piece be denoted by c. In that case, the control limits for c chart are used. The upper control limit (UCL) and lower control limit (LCL) for c chart is given by,

UCL = c-bar + 3 s.d

LCL = c-bar - 3 s.d

where c-bar = 1/n(€c) average number of defects

So, c-bar is computed as,

c-bar = (1+1+2+3+1+0+2+0+0+1+3+1+0+ 1+2+0 +0+2 + 1+2 +3+ 2+1+1) /24 = 30/24 = 1.25

put the data in Excel sheet and calculate required limits using Excel Commands.

Excel Command for Average, c-bar

=Average( Range of numbers who's average you want to calculate )

In my excel sheet, = Average ( H2 : H25)

= 1.25

Excel Command for Standard deviations,

=STDEV( numbers who's standard deviations you want to calculate)

here, =STD(H2:H25) , output= 0.98907

UCL = 1.25 + 3( 0.98907)

= 4.217

LCL = 1.25 - 3( 0.98907)

= - 1.717

but defective item can never be negative so, Lower control limit is Zero.

Hence, for the given data, the upper control limit for the number of defects is 4 and the lower control limit is 0.

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The change in volume of the sphere or best approximation to the increase in the volume of the sphere is 3.6π cubic inches.

In simple words, when a sphere is placed in the space or three-dimension space, then the total space acquired by the sphere generally shows the volume of the sphere. The mathematical formula needed to determine the volume of a sphere is shown below,

[tex]V = \frac{4}{3} \pi r^3[/tex]

Given- The initial radius of the sphere is r1 = 3 inch and the final radius of the sphere is r2 = 3.1 inch.

Now , the change in the radius of the sphere is [tex]dR =[/tex][tex]r2 - r1 =\\[/tex] 3.1 - 3 = 0.1 inch.

The mathematical expression of the volume of a sphere is shown below,

[tex]V = \frac{4}{3}\pi r^3[/tex]   (here, [tex]r[/tex] is the radius of sphere)

Differentiate the above shown expression with respect to radius ([tex]r[/tex]) as shown below,

[tex]\frac{d}{dr} (V) = \frac{d}{dr} (\frac{4}{3} ) \pi r^3[/tex]

∴

dV/dR = 4/3 π × d/dR(r^3)

          = 4/3 π × 3r^2

dV/dR = 4π r^2

∴ dV = 4π r^2 dR            ................(1)

Substitute all values in equation (1) and calculate the approximation as shown below,

dV =  4π r^2 dR

     =  4π (3)^2 × (0.1)

∴ dV = 3.6π cubic inches.

The change in volume of the sphere or best approximation to the increase in the volume of the sphere is 3.6π cubic inches.

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Option C, The research situation that would be most likely to use an independent-measures design is to compare the mathematics skills of 9th-grade boys versus 9th-grade girls.

An independent-measures design is used when there are two or more groups that are being compared and each individual only belongs to one group. In this research situation, the two groups being compared are 9th-grade boys and 9th-grade girls. Each individual in the study belongs to one group or the other, and they are not part of both groups. This makes it an ideal situation for using an independent-measures design.

Option a) examine the development of vocabulary as a group of children mature from age 2 to age 3 would likely use a within-subjects design, as the same group of children is being measured at two different time points.

Option b) Examining the long-term effectiveness of a stop-smoking treatment by interviewing subjects 2 months and 7 months after the treatment ends would also likely use a within-subjects design, as the same group of subjects is being measured at two different time points.

Option d) compare the blood-pressure readings before medication and after medication for a group of patients with high blood pressure would likely use a repeated-measures design, as the same group of patients is being measured at two different time points and the order of the measurements (before or after medication) is being controlled.

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Let x be the length of the shorter wall of the rectangular room and y be the length of the longer wall of the rectangular room. Since the longer wall is 7 ft longer than the shorter wall, we can write the equation y = x + 7. We can also express the diagonal of the rectangular room using the Pythagorean theorem as x^2 + y^2 = 13^2. We can solve for x and y by substituting the equation y = x + 7 into the equation x^2 + y^2 = 13^2 and then solving for x. Doing this, we get x^2 + (x + 7)^2 = 13^2. Expanding the square on the right side of the equation and then rearranging the terms, we get x^2 + 2x^2 + 14x + 49 = 169. Combining like terms, we get 3x^2 + 14x - 120 = 0. This quadratic equation can be factored as (x - 8)(3x + 15) = 0. Since the length of a side of a rectangle must be positive, we can ignore the solution x = -15/3. So, the length of the shorter wall of the rectangular room is x = 8 ft. The length of the longer wall can be found by substituting this value into the equation y = x + 7, giving us y = 8 + 7 = 15 ft. Therefore, the dimensions of the rectangular room are 8 ft by 15 ft.

The value of tangent AB as calculated from the given data is 14.4.

The tangents to the circle as given are PA and PB.

Let the point of intersection of PO and AB be X.

PA = 12  

OA = 9

As given that the triangle is right angled we can use hypotenuse theorem to calculate PO,

PO = √ PA²  + OA²

= √ 144 + 81

= 15

Now , the given triangles are similar because,

 AX / AO  =  PA / PO

 AX / 9  =  12 / 15

Therefore, the value of X is,

X  =  7.2

Now we know that AB's midpoint is X,

Hence , AB = 2X = 14.4

Tangent is known as a point which passes through a circle only at one single point A circle can have infinite tangents.

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The expected value E(X) is 7 and the value of variance of X is 5.8333.

Two dice are rolled. Let X be the random variable measuring the sum of the two numbers rolled.

a) The probability of mass function is obtained below:

The possible outcomes in each of the dice are 1 to 6. Therefore, the possible outcomes when two dice is rolled is 36

The sample space, s for fair dice (red die and blue die) is given below:

N(s) = 36

From the given information, two dice are rolled let X be the random variable measuring the sum of the two numbers rolled.

b) The expected value is calculated below:

The probability mass function of X is,

The required mean is,

E(x) = ∑xP(X=x)

[tex]=[2[/tex]×[tex]\frac{1}{36}+3[/tex]×[tex]\frac{2}{36}[/tex]+....+11×[tex]\frac{2}{36}+12[/tex]×[tex](\frac{1}{36} )[/tex]]

=[0.0556+0.1667+0.3333+0.5556+0.8333+1.1111+1.000+0.8333+0.6111+0.3333

=7

C) The variance V(X) is calculated below:

The probability mass function of X is,

The required variance is,

V(x)= ∑[tex](X-x)^{2}P(x) = 5.8333[/tex]

Therefore, the expected value E(X) is 7 and the value of variance of X is 5.8333.

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Answer:

5 years = 2,700.00

25 years = 30,600.00

Step-by-step explanation:

Hey, these are difficult, I know.

I'll explain how you solve them.

In this case, $90,000.00 needs to first be divided by 1,000.

That gives us 90.

You then multiply 90 by 30, which is the 5 yr cash value.

That gives you 2,700.00.

Since you have 90, you next need to multiply that by 340 (25 yr cash value), which gives you 30,600.00.

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The fifth percentile of the standard normal distribution is -1.64.

In the given question, we have to determine the fifth percentile of the standard normal distribution as a decimal rounded to the nearest hundredth.

The standard normal distribution, commonly known as the z-distribution, is a unique type of normal distribution in which the mean and standard deviation are both equal to 1. Any normal distribution's values can be transformed into z scores to normalise it. Z score shows the amount of standard deviations from the mean that each consists of numbers.

Since, X is normally distributed.

P(X< c) = 0.05

Using Z table

c = -1.645

Fifth percentile = -1.64

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Answer: slope = -1/4 y-intercept = -1/2

Step-by-step explanation:

Given an equation in the form y = mx + b, m = slope and b = y-intercept.

We can rearrange y= -1/2 - 1/4x into that form.

y = -1/4x - 1/2

We can now obviously see that -1/4x is the slope, and -1/2 is the y-intercept.

For the normal distribution of helium porosity (in percentage) of coal samples,

a) 95% CI for the true average porosity of a certain seam is equals to ( 4.56 , 5.14 ).

b) A 98% CI for true average porosity of another seam based on 14 specimens is equals to (4.11 , 5.01).

c) Sample size with width of confidence interval 0.42 is 45.

d) Necessary Sample size to estimate true average porosity to within 0.24 is 60.

We have Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed.

Standard deviations, σ = 0.72

a)Average porosity of specimens , X-bar = 4.85

Sample size, n = 23

Confidence level , alpha = 95% = 0.95

From Normal Distribution table

Z for 95% Confidence Interval = 1.96

so,95% Confidence interval = (x-bar - Z× σ/sqrt(n) , x-bar + Z×σ/sqrt(n) )

plugging all known values in above formula,

= ( 4.85-1.96×0.72/sqrt(23) , 4.85+1.96×0.72/sqrt(23) )

So, 95% Confidence interval = ( 4.56 , 5.14 )

b)Now, Sample size, n = 14

sample mean, X-bar = 4.56

from normal distribution Z- table

Z-score for 98% CI is equal to 2.33

so,98% CI = (x-bar-Z×σ/sqrt(n) , x-bar+Z×σ/sqrt(n) )

= ( 4.56-2.33×0.72/√14 , 4.56+2.33×0.72/√14 )

98% CI = ( 4.11 , 5.01 )

c)Using the normal Distribution Z- table

Z for 95% CI = 1.96

Width of confidence interval = 0.42

we have to determine the sample size, n .

Using formula, Width = 2×Z×σ/sqrt(n)

=> n = (2×Z×sd/width)²

=> n= (2×1.96×0.72/0.42)²

=> n = 45.15

=> n = 45 ( whole number)

d) From normal distribution table

Z for 99% CI = 2.58

Margin of error, ME = 0.24

margin of error = ME = Z×σ/sqrt(n)

=> n = (Z× σ/ME)²

=> n= (2.58×0.72/0.24)²

=> n = 59.9

=> n = 60

Hence, required sample size is 60.

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