Yezenia is filling up her outdoor swimming pool to get ready for summer time. She leaves the garden hose running for 1000 minutes in order to fill the pool, which has a volume of 15,460 gallons. What is the approximate rate at which water leaves the garden hose?

Answer:

Select all possible values for x in the equation.

x cubed=375.

5*the cubed root of 3

the cubed root of 375

75*the cubed root of 5

125*the cubed root of 3

I am trying to do a practice test to prepare for my real test tomorrow and I don't understand the question. Can anyone help explain it plz any help would be great.

Step-by-step explanation:

a) the reading on the thermometer after 3 more minutes will be -3 degrees Celsius.

b) the thermometer will read 6 degrees Celsius after 1.5 minutes.

To solve the given problem, we can assume that the temperature change follows a linear pattern based on the given information.

(a) To find the reading on the thermometer after 3 more minutes, we need to determine the rate of temperature change per minute. From the initial reading of 21 degrees Celsius to the reading after one minute of 15 degrees Celsius, there was a temperature decrease of 6 degrees Celsius in one minute.

Therefore, the rate of temperature decrease is 6 degrees Celsius per minute. If this rate remains constant, after 3 more minutes, the thermometer will show a further temperature decrease of:

3 minutes * 6 degrees Celsius per minute = 18 degrees Celsius

Thus, the reading on the thermometer after 3 more minutes will be 15 degrees Celsius - 18 degrees Celsius = -3 degrees Celsius.

(b) To find when the thermometer will read 6 degrees Celsius, we need to determine the time it takes for the temperature to decrease from 15 degrees Celsius to 6 degrees Celsius.

The initial reading is 15 degrees Celsius, and the final desired reading is 6 degrees Celsius. Therefore, we need to calculate the time it takes for a temperature decrease of:

15 degrees Celsius - 6 degrees Celsius = 9 degrees Celsius

Since the rate of temperature decrease is 6 degrees Celsius per minute, we can set up the equation:

9 degrees Celsius = 6 degrees Celsius per minute * t minutes

Solving for t (the time it takes to reach 6 degrees Celsius):

t = 9 degrees Celsius / 6 degrees Celsius per minute = 1.5 minutes

Therefore, the thermometer will read 6 degrees Celsius after 1.5 minutes.

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The value of limits after using limit laws is [tex]$\lim_{x \to 2} \frac{h(x) \cdot g(x)}{f(x)} = -\frac{28}{3}$.[/tex]

What are Limit Laws?

Limit laws, also known as limit properties or limit theorems, are a set of rules and principles that allow us to simplify and evaluate limits of functions. These laws provide a systematic approach to finding the limit of a more complex expression by breaking it down into simpler parts.

Given:

[tex]\lim_{x \to 2} f(x) &= -3 \\\lim_{x \to 2} g(x) &= 4 \\\lim_{x \to 2} h(x) &= 7\end{align*}\textbf{(a) Calculate} $\lim_{x \to 2} (f(x) - 2g(x))$:[/tex]

Using the limit laws, we can split the expression and apply the limit laws individually:

[tex]\lim_{x \to 2} (f(x) - 2g(x)) &= \lim_{x \to 2} f(x) - \lim_{x \to 2} (2g(x)) \\&= \lim_{x \to 2} f(x) - 2 \lim_{x \to 2} g(x) \\&= (-3) - 2(4) \\&= -3 - 8 \\&= -11[/tex]

Therefore,[tex]$\lim_{x \to 2} (f(x) - 2g(x)) = -11$.[/tex]

[tex]\textbf{(b) Calculate} $\lim_{x \to 2} (h(x))^2$:[/tex]

Again, using the limit laws, we can apply the limit to the expression:

[tex]\lim_{x \to 2} (h(x))^2 &= \left(\lim_{x \to 2} h(x)\right)^2 \\&= (7)^2 \\&= 49[/tex]

Therefore,

[tex]\lim_{x \to 2} (h(x))^2 = 49$.\textbf{\\\\(c) Calculate} $\lim_{x \to 2} \frac{h(x) \cdot g(x)}{f(x)}$:[/tex]

Applying the limit laws, we can evaluate the limit as follows:

[tex]\lim_{x \to 2} \frac{h(x) \cdot g(x)}{f(x)} &= \frac{\lim_{x \to 2} h(x) \cdot \lim_{x \to 2} g(x)}{\lim_{x \to 2} f(x)} \\\\&= \frac{7 \cdot 4}{-3}\\ \\&= \frac{28}{-3}[/tex]

Therefore,[tex]$\lim_{x \to 2} \frac{h(x) \cdot g(x)}{f(x)} = -\frac{28}{3}$.[/tex]

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The probability of three people sharing the same birthdays is approximately [tex]0.0000075[/tex] or [tex]0.00075[/tex]%, and there are three different pairs of people when there are three humans.

In conclusion, the probability of three people sharing the same birthdays is extremely low (approximately [tex]0.0000075 \ or \ 0.00075\%[/tex]), and when there are three humans, there exist three different pairs of people.

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a) The function f is not injective but is surjective.

b) The function f is injective and surjective.

c) The function f is not injective but is surjective.

d) The function f is not injective and not surjective.

a) The function f maps four elements from the domain {a, b, c, d} to five elements in the codomain {1, 2, 3, 4, 5}. Since there are more elements in the codomain than the domain, f cannot be injective. However, since every element in the codomain is mapped to by at least one element in the domain, f is surjective.

b) The function f(x) = x + 1 is a linear function that maps every real number to a unique real number. Hence, f is injective. Additionally, for every real number y, there exists x = y - 1 such that f(x) = y, meaning f is surjective.

c) The function f(m, n) = m + n maps pairs of integers from the domain Z × Z to integers in the codomain Z. Since there are infinitely many pairs that can result in the same sum, f cannot be injective. However, for every integer in the codomain, there exists at least one pair of integers in the domain whose sum is equal to it, making f surjective.

d) The function f(m, n) = m^2 + n^2 maps pairs of integers from the domain Z × Z to integers in the codomain Z. Since different pairs of integers can have the same sum of squares, f is not injective. Furthermore, there are integers in the codomain that cannot be obtained as a sum of squares, making f not surjective.

In summary, the injectivity and surjectivity of the given functions are as follows: a) not injective, surjective; b) injective, surjective; c) not injective, surjective; d) not injective, not surjective.

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Answer : a) a random variable X can only take the six values (1,2,3,4,5, and 6 ). If each value has equal probability, then p(1) = p(2) = p(3) = p(4) = p(5) = p(6) = 1/6

b) The pmf of the random variable X is:p(0) = 1/9p(1) = 2/9p(2) = 2/9p(3) = 4/9

Explanation :

A probability mass function is a function that gives the probability that a discrete random variable is exactly equal to some value.[1] Sometimes it is also known as the discrete density function.

The probability mass function is often the primary means of defining a discrete probability distribution, and such functions exist for either scalar or multivariate random variables whose domain is discrete.

a) If each value has equal probability, the pmf of the random variable X which can only take the six values (1,2,3,4,5, and 6) is : p(1) = p(2) = p(3) = p(4) = p(5) = p(6) = 1/6

b)If the probabilities of X(0,1,2, and 3) are 1/9,2/9,2/9, and 4/9. The pmf of the random variable X is:p(0) = 1/9p(1) = 2/9p(2) = 2/9p(3) = 4/9

The sum of these probabilities is:p(0) + p(1) + p(2) + p(3) = 1/9 + 2/9 + 2/9 + 4/9 = 9/9 = 1

So, the pmf is defined for all X.

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Step-by-step explanation:

The axis of symmetry: is the line that makes the parabola split in exactly half and lines up with the vertex. For that parabola x=1 is the line of symetry.

The vertex is where the minimum of the graph is, on this graph you can eyeball it to be (1,-9)

The x-intercept is where y is 0 so that's where the lines intersex with the x-axis. (-2,0) and (4,0)

The y-intercept of the function is where x is 0 and where the parabola intersects with the y-axis. On this graph it would be (0,-8)

Hope that helps :)

Answer:

D

Step-by-step explanation:

D

Answer:

The answer is 7 (A)

Step-by-step explanation:

Answer:

D. The length is the same as line MN.

Step-by-step explanation:

Even if the line is moved any one bit, the length of the line will not change. So as a result, A, B & C are already out. That leaves D as our answer.

Given that a random-sample of 250 people is surveyed to determine if they own a tablet, where 98 people own a tablet.

We have to find a confidence interval estimate for the true proportion of people who own tablets using a 95% confidence-level.

The formula to compute confidence interval estimate is given by;

[tex]CI = p \pm Z_{\frac{\alpha}{2}}\sqrt{\frac{p(1-p)}{n}}[/tex]

Where;[tex]p[/tex] = Sample proportion[tex]Z_{\frac{\alpha}{2}}[/tex] = Critical value of Z at [tex]\frac{\alpha}{2}[/tex][tex]n[/tex] = Sample size

From the given data,Sample proportion, [tex]p = \frac{98}{250} = 0.392[/tex]

Level of Confidence, [tex]C= 95%[/tex]

As level of significance [tex]\alpha = (1-C) = 0.05[/tex]So, [tex]\frac{\alpha}{2} = \frac{0.05}{2} = 0.025[/tex]

Sample size, [tex]n = 250[/tex]

Now, we need to find the critical value of [tex]Z_{0.025}[/tex] such that the area to its right in the z-distribution is 0.025.Z-table shows the values of Z for given probabilities.

The closest value to 0.025 is 1.96. So, we can take [tex]Z_{0.025} = 1.96[/tex].

Therefore, the confidence interval estimate for the true proportion of people who own tablets using a 95% confidence level is given as;[tex]CI = 0.392 \pm 1.96\sqrt{\frac{0.392(1-0.392)}{250}}[/tex][tex]\Rightarrow CI = 0.392 \pm 0.067[/tex]

So, the lower limit of the interval is obtained as;

[tex]0.392 - 0.067 = 0.325[/tex]

And the upper limit of the interval is obtained as;

[tex]0.392 + 0.067 = 0.459[/tex]

Therefore, with 95% confidence, we say that the proportion of people who own tablets is between 32.5% and 45.9%.

The correct option is (A).

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The level of measurement used in the operationalization of extracurricular participation is categorical/nominal.

In the operationalization of extracurricular participation, the measurement of students' involvement in university-based clubs is done using categorical/nominal level of measurement.

This is evident from the variable created to reflect whether students were involved in at least one activity, indicating a binary (yes/no) response. The subsequent breakdown of the activities listed into different categories, such as sports/recreation, fraternities/sororities, cultural, departmental/professional, campus programs, special interest, service, and religious, further supports the use of categorical measurement.

Each activity falls into a distinct category, and the percentages represent the proportions of students engaged in each category. Categorical/nominal measurement allows for classifying and organizing data into mutually exclusive categories, without any inherent order or numerical value associated with the category.

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The value of dy/dt for the given equation and values is -6.

To evaluate dy/dt, we can differentiate the given equation with respect to t using the chain rule. Starting with the equation 4xy - 3x + 4y^3 = -76, we differentiate both sides with respect to t.

Differentiating each term separately, we get:

(d/dt)(4xy) - (d/dt)(3x) + (d/dt)(4y^3) = 0

Using the chain rule, we can rewrite this as:

4(dy/dt)(x) + 4x(dy/dt) - 3(dx/dt) + 12y^2(dy/dt) = 0

Substituting the given values dx/dt = -8, x = 4, and y = -2, we have:

4(dy/dt)(4) + 4(4)(dy/dt) - 3(-8) + 12(-2)^2(dy/dt) = 0

Simplifying the equation, we get:

16(dy/dt) + 16(dy/dt) + 24 + 48(dy/dt) = 0

80(dy/dt) = -24

(dy/dt) = -24/80

(dy/dt) = -3/10

(dy/dt) = -0.3

Therefore, dy/dt evaluates to -0.3.

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Answer:

answer is 1

Step-by-step explanation:

(a) The probability that a randomly chosen wedge of cheddar cheese is greater than 10.14 is found using the standard normal distribution as follows:

P(Z > z) = P(Z > (10.14 - µ)/σ)

= P(Z > (10.14 - 10.2)/0.2)

≈ 0.3085.

Therefore, the probability is approximately 0.3085.

(b) If a sample of 16 is randomly chosen, then the distribution of the sample mean weight is approximately normal with a mean of 10.2 ounces and a standard deviation of σ/√n,

Where n = 16.

The sample standard deviation is given by σ = 0.2, so the standard deviation of the sample mean weight is:

σ/√n = 0.2/√16

= 0.05.

Therefore, the distribution of the sample mean weight is approximately normal with a mean of 10.2 ounces and a standard deviation of 0.05 ounces.

(c) The probability that the sample mean weight of this sample of 16 is less than 10.14 is found using the standard normal distribution as follows:

P(Z < z) = P(Z < (10.14 - µ)/(σ/√n))

= P(Z < (10.14 - 10.2)/(0.2/√16))

≈ P(Z < -1.6)

≈ 0.0548.

Therefore, the probability is approximately 0.0548.

(d) The probability that the sample mean weight of this sample of 16 is greater than 10.14 is found using the standard normal distribution as follows:

P(Z > z) = P(Z > (10.14 - µ)/(σ/√n))

= P(Z > (10.14 - 10.2)/(0.2/√16))

≈ P(Z > -1.6)

≈ 0.9452.

Therefore, the probability is approximately 0.9452.

(e) The probability that the sample mean weight of this sample of 16 is between 10.14 and 10.3 is found

Using the standard normal distribution as follows:

P(a < Z < b) = P((a - µ)/(σ/√n) < Z < (b - µ)/(σ/√n))

= P((10.14 - 10.2)/(0.2/√16) < Z < (10.3 - 10.2)/(0.2/√16))

≈ P(-1.6 < Z < 2)

≈ 0.9452 - 0.0548

= 0.8904.

Therefore, the probability is approximately 0.8904.

(f) Let x be the average weight of a sample of these 16 cheese wedges that is below some value z.

Then, the probability that x is less than z is 0.07.

Using the standard normal distribution, we can find the z-score such that

P(Z < z) = 0.07 as follows:

z = inv Norm(0.07)

≈ -1.4758.

Therefore, the average weight of a sample of these 16 cheese wedges that is below the value z is:

x = µ + z(σ/√n)

= 10.2 + (-1.4758)(0.2/√16)

≈ 10.0625.

Therefore, there is only a 7% chance that the average weight of a sample of these 16 cheese wedges will be below 10.0625.

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Answer: 11

Step-by-step explanation:

Sooooo all I did was take the formula for the area of a triangle ( A= 1/2(b)(h) 0 and plug In the values. So 'b' would be 4 and 'h' would be 5.5. I assumed this rectangle had equal widths and equal heights.

Answer:

125

Step-by-step explanation:

5 to the 3rd power means 5³

5³ = 5x5x5

= 125

The range of the repair costs for the four car crashes is $5452.54 - $232 = $5220.54. The sample variance of the repair costs is $4,898,414.69, and the sample standard deviation is $2,214.17.

The range of the repair costs for the four car crashes is the difference between the highest and lowest cost, resulting in a range of $5220.54. This indicates the variability in the repair costs. The sample variance is a measure of the average squared deviation from the mean, calculated to be $4,898,414.69. It shows the dispersion of the repair costs from the average. The sample standard deviation is the square root of the variance, amounting to $2,214.17. It provides a measure of how spread out the repair costs are, with a higher value indicating greater variability.

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By supposing that 7₁, 72 are linearly dependent in a vector space V. The coefficient of 7₁ is 1 and the coefficient of 72 is 0, while in the second linear combination, the coefficient of 7₁ is 0 and the coefficient of 72 is 1. Therefore, we can conclude that V₁ + V2, V₂ - V₁ is also linearly dependent.

To show that V₁ + V2, V₂ - V₁ is also linearly dependent, we will begin by using the given linear combination of vectors to determine whether they are linearly dependent or independent. Suppose that 7₁, and 72 are linearly dependent in a vector space V.

Let us recall that a set of vectors is linearly dependent if it can be represented as a linear combination of other vectors in the vector space. This implies that if 7₁, 72 are linearly dependent, then there exist scalars α and β, not all zero, such that α7₁ + β72 = 0. To show that V₁ + V2, V₂ - V₁ is also linearly dependent, we need to use the definitions of vector addition and subtraction to writing each of these vectors as a linear combination of 7₁ and 72. Let's begin with V₁ + V2.

Using the definition of vector addition, we have

V₁ + V2 = 1 · 7₁ + 1 · 72 = 7₁ + 72.

Similarly, using the definition of vector subtraction, we have

V₂ - V₁ = -1 · 7₁ + 1 · 72 = -7₁ + 72.

Now we can write V₁ + V2, V₂ - V₁ as linear combinations of 7₁ and 72:

V₁ + V2 = 1 · (7₁ + 72) + 0 · (-7₁ + 72)V₂ - V₁

= 0 · (7₁ + 72) + 1 · (-7₁ + 72)

Notice that the coefficients in each linear combination are not all zero. We can say that V₁ + V2, V₂ - V₁ is linearly dependent.

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Answer:

oblique

Step-by-step explanation:

Answer:

84500 is the correct answer

Answer:

40182 delta math

Step-by-step explanation:

Answer:

[tex]f(t) = 1450(1.025)^{t}[/tex]

Step-by-step explanation:

Given

[tex]P =1450[/tex] -- principal

[tex]r = 2.5\%[/tex] --- rate

[tex]n = 1[/tex] --- compounded once a year

Required

Determine the function for compound interest

Compound interest f(t) is calculated as:

[tex]f(t) =P(1 + r/n)^{nt[/tex]

So, we have:

[tex]f(t) = 1450(1 + 2.5\%/1)^{1 * t}[/tex]

[tex]f(t) = 1450(1 + 2.5\%)^{t}[/tex]

[tex]f(t) = 1450(1 + 0.025)^{t}[/tex]

[tex]f(t) = 1450(1.025)^{t}[/tex]

Main Answer: The base of C' is {0110, 1001, 1100, 0011}.

Supporting Explanation: In a communication system, codification and decodification are used to encode and decode messages. C is the code for the message, where C={0000, 1100, 1010, 0110, 0101, 0011, 1001, 1111}. The code is a binary code since F=Z2. C' is the dual code of C. The codewords in C' are orthogonal to those in C. A basis for C' can be determined by finding a generator matrix for C'. Thus, the generator matrix for C is the parity check matrix for C'. A generator matrix for C is given as, G = [I | P] where P is the parity check matrix. The parity check matrix for C can be determined as, P = [-AT | Im-k]. Therefore, P = [0101; 1010; 1111].The rows of C' correspond to the columns of P. Thus, a basis for C' is {0110, 1001, 1100, 0011}.

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answer:

3 plants are left !

step-by-step explanation:

hi there!

first lets use our times table to see which number gets use closest to 35 when it is multiplied by 4

4, 8, 12, 16, 20, 24, 28, 32,

4 * 8

= 32

then you can just subtract 35 by 32 to get you final answer of

35 - 32

= 3 plants are left over

( i hope i did the problem right since the wording of it was kinda hard to understand it )

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i made a table with each number being the plant! just in case your a visual learner like me!

but i hope this help you and if i made anything confusing or if i made a mistake please let me know! but other then that i hope you have a good rest of your day :)

There are 3 plants left.

What is Division method?

Division method is used to distributing a group of things into equal parts. Division is just opposite of multiplications.

For example, dividing 20 by 2 means splitting 20 into 2 equal groups of 10.

Given that;

Total number of plants = 35

And, You want to split them equally between 4 sections of your garden.

Now,

To find the left plants after splitting into 4 sections, we can divide and find the remainder as;

The number of left plants = 35 ÷ 4

 4 ) 35 ( 8

    - 32

    -------

         3

Clearly, The remainder = 3

Thus, There are 3 plants left.

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Answer:

24 units by 15 units

Step-by-step explanation: To find how many units the length and width are, divide each by 5:

120/5 = 24

75/5= 15

For every 5 feet, there is 1 unit .

Answer:

$27.50

Step-by-step explanation:

Hope this helps and have a great day!!!!

Step-by-step explanation:

Since his pay is increased by 10 percent, then you multiply 10% by 25 and then add that to 25.

25+10/100(25)=55/2

27.5

Hope that helps :)

Answer:

122

Step-by-step explanation:

180 - (43 + 15)

Answer:

The answer is 2.86

Step-by-step explanation:

you just have to divide 43 by 15

Answer:

x=20

Step-by-step explanation:

Hello There!

Remember the exterior angle of a triangle rule:

An exterior angle of a triangle is equal to the sum of the opposite interior angles

Knowing this, we can create an equation to solve for x

exterior angle (100) = sum of opposite interior angles (3x+2x)

100 = 2x+3x

now we solve for x

step 1 combine like terms

2x+3x=5x

now we have 100=5x

step 2 divide each side by 5

5x/5=x

100/5=20

we're left with x = 20

Answer:

With their steel hoofs, their long legs, their stag-like muscles, their thick skins, their powerful horns, they could walk the roughest ground, cross the widest deserts, climb the highest mountains, swim the widest rivers, fight off the fiercest bands of wolves, endure hunger, cold, thirst and punishment as few beasts of the earth have ever shown themselves capable of enduring.