At 4:00 a.m., the outside temperature was -28°F. How much warmer does 1 pc
it have to get to reach 0°F?

Answer:

no.

Step-by-step explanation:

The fewest reported number of text messages sent, given the data is 62 (Option B)

The fewest number in a given data is simply defined as the smallest number in the data.

Considering the data given from the question, we can see that the smallest number in the data is 62

Thus, 62 is the fewest number in the data.

Therefore, the correct answer to the question is 62 (Option B)

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Complete question

The Box-and-Whisker Plot shown displays the number of text messages sent during the previous week. What is the fewest reported number of text messages sent? 62, 74, 82, 90, 99

A. 82

B. 62

C. 74

D. 90

Answer:

The answer is 11/24r - 8.

Step-by-step explanation:

Please give brainliest

a) The stem and leaf plot of the data is illustrated below.

b) Based on the stem-and-leaf plot, it appears likely that a coupon will "survive" beyond 2000 cycles since there are data points in that range.

First, we separate the data into stems and leaves. The stem represents the tens place, while the leaves represent the ones place. For example, if a data point is 1223, the stem would be 122 and the leaf would be 3. By organizing the data in this manner, we can easily determine the frequency of values and observe any patterns that emerge.

Now let's construct the stem-and-leaf plot for the given data:

Stem | Leaves

------+-------

7 | 06 58 85 90 98

8 | 45 65 85 85 88 90 90 93 94 94 98

9 | 10 55 69 69 90 90 90 90 90 93 94 94 94

10 | 15 20 23 23 26 30 33 40 42 42 50 58 60 62 62 74 85 90

11 | 02 02 06 09 20 20 22 23 26 35 52 52 52 52 55 67 90 90

12 | 03 20 20 22 23 33 33 58 58 69 90 90

13 | 15 35 42 46 58 70 82 98

14 | 02 35 42 81

15 | 18 23 35 40 42 52 67 74

16 | 42 60 42

17 | 06 30 50 81

18 | 20 22 50 82

19 | 30 90 92

20 | 10 30 50 70 82

21 | 00 00 02

22 | 33 33 58 58 90

23 | 33 33

24 | 20 22

25 | 33 65

For instance, a stem value of 8 with leaves 45 indicates two data points: 845 and 885. The frequency of each data point can be determined by counting the number of leaves associated with each stem.

Now, let's address the question of whether it appears likely for a coupon to "survive" beyond 2000 cycles. To answer this, we examine the stem-and-leaf plot. We observe that the stem value of 20 contains several leaves greater than 0, indicating that there are data points in the 2000s range. Additionally, the stem values of 21 and 22 also have leaves, suggesting the presence of data points beyond 2000 cycles.

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The points are (0,0) Minimum and (1/6, 1/18) is the saddle point.

What is the saddle point?

To solve the question we have:

f(x,y) = xy(1-2x-6y)

Therefore, f(x,y) = xy-2x²y-6xy²

Finding the partial derivative of the above gives:

fₓ = y - 4xy-6y²

= x - 2x²-12xy

Therefore, when the above equations are set to 0 we get:

y - 4xy-6y² = 0.......(1)

x - 2x²-12xy  = 0......(2)

Dividing equations (1) and (2) by y and x respectively give:

1 - 4x - 6y = 0

1 - 2x - 12y = 0  

Solving gives:

x = 0.167 = 1/6

y = 0.056 = 1/18

When we place x in equation (1) we have:

y - 4(1/6)y-6y² = 0

y = 0 or y = 1/18

Similarly, x = 0 or 1/6

Therefore, the four critical points are:

(0,0) (1/6,1/18)

When x = 0 and y = 0

f(x,y) = 0 Hence this is a minimum

When x = 1/6 and y = 1/18

f (x,y) = 1/324 Which is a saddle point.

Therefore, the points are (0,0) Minimum, and (1/6, 1/18) is the saddle point.

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The question you are looking for is here:

(1 point) The function f(x,y)=xy(1−2x−6y) has 4 critical points. List them and select the type of critical point. Points should be entered as ordered pairs and listed in increasing lexicographic order. By that, we mean that (x,y) comes before (z,w) if x

Answer:

The bottom right table

Step-by-step explanation:

Answer:

----------------------

According to the diagram we have:

1) Angles w and 138° form a linear pair, hence:

2) Angles 19°, x and w form a right angle, hence:

Answer:5,-3

Step-by-step explanation:one easy way is to start on the y axis and count the spaces towards “b” and so it would be 5 spaces to the left of the y axis so you are flipping it so you go the other way 5 spaces like a mirror essentially

The mean number of individuals with 20/20 vision is 23.

To find the mean number of individuals with 20/20 vision, we can use the formula for the expected value of a binomial distribution. In this case, the probability of an individual having 20/20 vision is p = 0.33, and the number of trials (i.e. individuals selected) is n = 70.
The formula for the expected value of a binomial distribution is:
E(X) = np
Substituting in our values, we get:
E(X) = 70 x 0.33
E(X) = 23.1
So, the mean number of individuals with 20/20 vision out of 70 selected at random from the population is approximately 23.1. However, since we can't have a fraction of a person, we should round our answer to the nearest whole number.
Therefore, the mean number of individuals with 20/20 vision is 23.

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AnswerAnswer:

1. no

2. no

4. no

6. yes, none of the numbers repeat

8. yes, none of the numbers repeat

Step-by-step explanation:

It has been a while since I did this so  not exaI'm not exactly sure about the rest of them.

Answer:

It has no solution.

Answer:

Actually i think it does have a solution, if it does it would be a=1/4, 3

If i'm wrong i am very sorry..... So, if that not right then your answer is that it has no solution.....

Stay safe and have a Merry Christmas!!!!!!! :D

Answer:

x = 4 ; a = 15 ; b= -2

Step-by-step explanation:

Answer: $1460

Step-by-step explanation:

He will add $40 for 24 months since there are 24 months in two years. 40 x 24 = 960 so you should add 960 and the original amount of 500. This ends up with $1460

Answer:

$1460

Step-by-step explanation:

So he already has $500 in his piggy.

There are 24 months in 2 years.

So, we would multiply 24 by $40

That would be $960

Add the $500 to $960.

$1460

Hope I helped :)

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Answer:

\(\huge\boxed{(x\cdot x^5):(x^2\cdot x)=x^3}\)

Step-by-step explanation:

\((x\cdot x^5):(x^2\cdot x)\qquad|\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=x^{1+5}:x^{2+1}=x^6:x^3\qquad|\text{use}\ a^n:a^m=a^{n-m}\\\\=x^{6-3}=x^3\)

Answer:

X × X^5 ÷ X^2 ×X

X^3

because the same variable can be cancelled.

The derivatives of the functions are :

(a) 4(x² + eˣ - ln x)³ (2x + eˣ - 1/x)

(b) -2/x³ - 1

(a) To calculate the derivative of the function f(x) = (x^2 + e^x - ln x)^4, we can use the chain rule and the power rule.

First, we can take the derivative of the function inside the parentheses using the sum and difference rule:

f'(x) = 4(x^2 + e^x - ln x)^3 (2x + e^x - 1/x)

Next, we can simplify this expression by multiplying the two factors together and raising the entire expression to the fourth power:

f'(x) = 4(x² + eˣ - ln x)^3 (2x + eˣ - 1/x)

Therefore, the derivative of the function

f(x) = (x² + eˣ - ln x)⁴ is

f'(x) = 4(x² + eˣ - ln x)³ (2x + eˣ - 1/x)

(b) To calculate the derivative of the function f(x) = (x(x² + 1)/x³) - x, we can use the quotient rule and the product rule.

First, we can simplify the expression inside the parentheses by multiplying x through:

f(x) = (x³ + x)/x³ - x

Next, we can take the derivative of this expression using the product rule and the power rule:

f'(x) = [x³(3x² + 1) - (x³ + x)3x²]/ x⁶ - 1

Simplifying this expression, we get:

f'(x) = [3x⁵ + x³ - 3x⁵ - 3x³]/ x⁶ - 1

= -2/x³ - 1

Therefore, the derivative of the function

f(x) = x(x² + 1)/x³ - x is

f'(x)  = -2/x³ - 1

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a) We would expect approximately 0.2325 × 200 = 46.5 vehicles

to fail in their first 2 years. Since we cannot have a fraction of a vehicle

failing, we would expect around 46 or 47 vehicles to fail in their first 2

years.

b) The approximate probability that 50 or more vehicles fail in

the first 2 years is 0.0004.

c) The approximate probability that no more than 10 of the remaining

vehicles fail in their first 2 years, given that 30 have already failed, is 0.0002.

a) The average lifespan of the vehicle is 8 years, and it follows an

exponential distribution, which has a probability density function of

\(f(x) = (1/θ) \times e^(-x/θ)\),

where θ is the mean of the distribution. Thus, θ = 8.

The probability that a vehicle fails in the first 2 years can be found by

integrating the exponential probability density function from 0 to 2:

P(X ≤ 2) = ∫[0,2] f(x) dx = ∫[0,2] (1/8) × e^(-x/8) dx

Using a calculator or a table of integrals, we can find this probability to

be approximately 0.2325.

b) The number of vehicles that fail in the first 2 years follows a Poisson

distribution with parameter λ = 200 × P(X ≤ 2) = 200 × 0.2325 = 46.5.

The probability that 50 or more vehicles fail in the first 2 years can be

found using the Poisson distribution with parameter λ = 46.5:

P(X ≥ 50) = 1 - P(X < 50)

Using a Poisson distribution table or a calculator, we can find that P(X <

50) is approximately 0.9996.

Thus, P(X ≥ 50) = 1 - 0.9996 = 0.0004 (approximately).

c) Given that 30 vehicles have already failed in under 2 years, we need

to find the probability that no more than 10 of the remaining 170 vehicles

fail in their first 2 years.

The number of vehicles that fail in the first 2 years follows a Poisson

distribution with parameter λ = 170 × P(X ≤ 2) = 170 × 0.2325 = 39.525

(rounded to 40).

Thus, we need to find P(Y ≤ 10), where Y is a Poisson random variable

with parameter λ = 40.

Using a Poisson distribution table or a calculator, we can find that P(Y ≤

10) is approximately 0.0002.

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the data derived from a sample statistic will show that the two variables that are being researched are related or that the relationship is not large enough for the variables to be related is true

Current statistical studies are modern and have a wealth of analysis tools and software to understand the data. Statistics has a wide range of objectives, which aim not only to understand the variability of the data, but also its origin and motivation. For example, if an election poll shows a variation from a previous poll, statisticians will seek to understand the causes of these variability and not just present the variation in the data.

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a) To calculate the pressure at 12 hours of shut-in:

substitute the given values into the pressure buildup equation and solve for P(t=12).

b) Superposition in time is used in pressure buildup analysis by adding or summing the responses of multiple transient tests to analyze and interpret reservoir behavior and properties.

We have,

a) To calculate the pressure in the production well at 12 hours of a shut-in, we can use the equation for pressure transient analysis during shut-in periods, known as the pressure buildup equation:

P(t) = Pi + (Q / (4πKh)) * log((0.14ϕμCt(t + Δt)) / (Bo(ΔP + Δt)))

Where:

P(t) = Pressure at time t

Pi = Initial reservoir pressure

Q = Flow rate

K = Permeability

h = Reservoir thickness

ϕ = Porosity

μ = Viscosity

Ct = Total compressibility

t = Shut-in time (12 hours)

Δt = Time since the start of the flow period

Bo = Oil formation volume factor

ΔP = Pressure drop during the flow period

Given:

Pi = 3100 psi

Q = 200 STB/day

K = 45 md

h = 60 ft

ϕ = 15%

μ = 1.2 cp

Ct = 15x10^-6 psi^-1

Bo = 1.3 bbl/STB

t = 12 hours

Δt = 48 hours

ΔP = Pi - P(t=Δt) = Pi - (Q / (4πKh)) * log((0.14ϕμCt(Δt + Δt)) / (Bo(ΔP + Δt)))

Substituting the given values into the equation:

ΔP = 3100 - (200 / (4π * 45 * 60)) * log((0.14 * 0.15 * 1.2 * 15x\(10^{-6}\) * (48 + 48)) / (1.3 * (3100 - (200 / (4π * 45 * 60)) * log((0.14 * 0.15 * 1.2 * 15 x \(10^{-6}\) * (48 + 48)) / (1.3 * (0 + 48))))))

After evaluating the equation, we can find the pressure in the production well at 12 hours of shut-in.

b) Superposition in time is a principle used in pressure transient analysis to analyze and interpret pressure build-up tests.

It involves adding or superimposing the responses of multiple transient tests to simulate the pressure behavior of a reservoir.

The principle of superposition states that the response of a reservoir to a series of pressure changes is the sum of the individual responses to each change.

Superposition allows us to combine the information obtained from multiple tests and obtain a more comprehensive understanding of the reservoir's behavior and properties.

It is a powerful technique used in reservoir engineering to optimize production strategies and make informed decisions regarding reservoir management.

Thus,

a) To calculate the pressure at 12 hours of shut-in:

substitute the given values into the pressure buildup equation and solve for P(t=12).

b) Superposition in time is used in pressure buildup analysis by adding or summing the responses of multiple transient tests to analyze and interpret reservoir behavior and properties.

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Answer:

V = (14)(4)   V = 56cm³

Step-by-step explanation:

The Volume is Base times Height

We are given that the base is 7cm by 2cm.

That is 14 cm²  (The cubes each have volume of 1 cm³, so the volume of that bottom row is 14cm³, technically, but ignore that for now.)

To find the volume, multiply the base times the height.

V = 14cm² × 4cm

V = 56cm³

the four points become:

(1, 0)
(0.707, 0.707)
(-1, 1.57)
(-0.707, 2.36)

Note that the angle is given in radians and the distance is measured in units of the polar coordinate system.

The polar equation given is:

r = cos(2θ)

To find points on four rays, we can substitute different values of θ.

For θ = 0, we have:

r = cos(2(0)) = cos(0) = 1

So the point on the ray directed at θ = 0 with distance 1 is (1, 0).

For θ = π/4, we have:

r = cos(2(π/4)) = cos(π/2) = 0

So the point on the ray directed at θ = π/4 with distance 1 is (0, π/4).

For θ = π/2, we have:

r = cos(2(π/2)) = cos(π) = -1

So the point on the ray directed at θ = π/2 with distance 1 is (-1, π/2).

For θ = 3π/4, we have:

r = cos(2(3π/4)) = cos(3π/2) = 0

So the point on the ray directed at θ = 3π/4 with distance 1 is (0, 3π/4).

To simplify the answers, we can convert the points to rectangular coordinates using the formulas:

x = r cos(θ)
y = r sin(θ)

Using these formulas, the four points become:

(1, 0)
(0.707, 0.707)
(-1, 1.57)
(-0.707, 2.36)

Note that the angle is given in radians and the distance is measured in units of the polar coordinate system.
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The cylindrical coordinates of (−7, 7, 7) are (√98, -π/4, 7). cylindrical coordinate (−7, 7 3 , 1) is  (14, -π/3, 1).

We must switch from rectangular to cylindrical coordinates in the provided problem. (let r ≥ 0 and 0 ≤ θ ≤ 2π.)

A)(−7, 7, 7)

Given rectangular coordinates(x, y, z) =(−7, 7, 7)

The cylindrical coordinates are (r, θ, z)

As a result, we determine each value of r and θ  separately.

r = √x²+y²

r = √(-7)²+(7)²

r = √49+49

r = √98

θ = tan⁻¹ (y/x)

θ =tan⁻¹  (7/-7)

θ =tan⁻¹  (-1)

θ = -π/4

So cylindrical coordinate = (r, θ, z) = (√98, -π/4, 7)

B) (-7,7√3,1)

Given rectangular coordinates(x, y, z) = (-1,1,1)

The cylindrical coordinates are (r, θ, z)

As a result, we determine each value of r and θ  separately.

r = √x²+y²

r = √(-7)²(7√3)²

r = √49+147

r = √196

r = 14

θ = (y/x)

θ = (7√3/-7)

θ = (-√3)

θ = -π/3

So cylindrical coordinate = (r, θ, z) = (14, -π/3, 1)

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Answer: I think A

Step-by-step explanation:

Answer:

Step-by-step explanation:

a) 1200≥25x+45y

b) YES, because:

1200≥25(15)+45(15)

1200≥375+675

1200≥1050

In mathematics, we can define ratio as how many times one number contains another. In the other word, a ratio says how much of one thing there is compared to another thing. Example, there are ten apples and six lemons in a basket of fruit, then the ratio of apples to lemons is ten to six (10:6). We can also say that the ratio of apples to the total amount of fruit is 10:16.

Ratios making us easier to understand the case by allow us to measure and express quantities.

Case:

- Alex's Salad contains oil and pasta.

- The ratio of ounces oil to cups of pasta is 3:5

How many ounces of oil will alex need to make 64 cups of pasta?

Oil = \(\frac{3}{5}\) x 64 = 38,4

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a. To show that the model can be written as xt = δt + ∑1twk, we can use induction.

For t = 1:

x1 = δ + x0 + w1

= δ + 0 + w1

= δ + w1

So the base case holds.

Assume that the model holds for t = k, i.e., xk = δk + ∑1k wk.

For t = k+1:

xk+1 = δ + xk + wk+1

= δ + (δk + ∑1k wk) + wk+1

= δk+1 + ∑1k+1 wk

Therefore, by induction, the model can be written as xt = δt + ∑1twk.

v. To find the mean function of xt, we can take the expected value of both sides of the model equation:

E[xt] = E[δt + ∑1twk]

= E[δt] + E[∑1twk]

= δt + ∑1tE[wk]

= δt

So, the mean function of xt is given by μt = δt.

To find the autocovariance function of xt, we need to calculate Cov(xt, xs) for s ≠ t.

For s < t:

Cov(xt, xs) = Cov(δt + ∑1twk, δs + ∑1swk)

= Cov(δt, δs) + Cov(δt, ∑1swk) + Cov(∑1twk, δs) + Cov(∑1twk, ∑1swk)

= 0 + 0 + 0 + Cov(∑1twk, ∑1swk)

= Cov(∑1twk, ∑1swk)

= Cov(wt + ∑1t-1wk, ws + ∑1s-1wk)

= Cov(wt, ws) + Cov(wt, ∑1s-1wk) + Cov(∑1t-1wk, ws) + Cov(∑1t-1wk, ∑1s-1wk)

= 0 + 0 + 0 + Cov(∑1t-1wk, ∑1s-1wk)

Since wt and wk are white noise with mean zero and variance σw^2, their covariance is zero unless t = s. Therefore, we have:

Cov(xt, xs) = 0 for s < t

For s > t, the covariance remains zero as well, since we can use the same logic as above.

For s = t, we have:

Cov(xt, xs) = Cov(δt + ∑1twk, δt + ∑1twk)

= Cov(δt, δt) + Cov(δt, ∑1twk) + Cov(∑1twk, δt) + Cov(∑1twk, ∑1twk)

= Var(δt) + Cov(∑1twk, ∑1twk)

= Var(δt) + Var(∑1twk) + 2Cov(∑1twk, ∑1twk)

= Var(δt) + Var(∑1twk) + 2∑1t-1Cov(wk, wk)

= Var(δt) + Var(∑1twk) + 2∑1t-1σw^2

Since δ is a constant and wk is a white noise process, Var(∑1twk) = ∑1tVar(wk) = tσw^2.

Therefore, we have:

Cov(xt, xs) = Var(δt) + tσw^2 + 2∑1t-1σw^2

= Var(δt) + tσw^2 + 2σw^2(t-1)

= Var(δt) + 3tσw^2 - 2σw^2

Finally, we can conclude that the mean function of xt is μt = δt, and the autocovariance function of xt is Cov(xt, xs) = Var(δt) + 3tσw^2 - 2σw^2.

c. To determine if xt is stationary, we need to check if the mean function and autocovariance function are independent of time (t).

In this case, the mean function is μt = δt, which is dependent on time. Therefore, xt is not stationary.

Similarly, the autocovariance function Cov(xt, xs) = Var(δt) + 3tσw^2 - 2σw^2 depends on both t and s, so xt is not stationary.

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The exponential function that shows Riley's target mileage is y = 12(1.1)ˣ

An equation is an expression that shows how numbers and variables are linked together using mathematical operations such as addition, subtraction, multiplication and division.

The standard exponential equation is:

y = abˣ

where b is the rate of change and a is the initial value.

Let y represent Riley target mileage x weeks after she starts training.

She plans to run 12 miles in her first week of training and will increase her mileage by about 10% each week after. Hence:

a = 12; b = 100% + 10% = 110% = 1.1

The exponential function is y = 12(1.1)ˣ

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Approximately 13 percent  of the student body in the 9th grade don't play a sport.

To find the percentage of the student body in the 9th grade who don't play a sport, we need to calculate the number of 9th-grade students who don't play a sport and divide it by the total number of students, then multiply by 100 to get the percentage.

From the table, we can see that the number of 9th-grade students who don't play a sport is 32. The total number of students is 240.

Percentage = (Number of 9th-grade students who don't play a sport / Total number of students) * 100

= (32 / 240) * 100

= 0.133333 * 100

≈ 13.33

Rounded to the nearest whole number, the percentage is 13.

Therefore, approximately 13% of the student body in the 9th grade don't play a sport.

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Answer: 15 is a 25% increase of 12.

Step-by-step explanation: Difference of 12 and 15 = |12 - 15|/((12 + 15)/2) = 3/13.5 = 0.22222222222222 = 22.222222222222%

Answer:

13/12

Step-by-step explanation:

Answer:

12/13

Step-by-step explanation:

remember it's adj/hyp so it would be 36/39. When simplified it becomes 12/13.