What is 3(x - 1) = 2x + 9

Answer:

2x - 4 = 14

When x = 9

Step-by-step explanation:

Use PEMDAS order of operations to solve:

2(9) - 4

= 18 - 4

= 14

Answer:

\(\displaystyle \frac{d}{dx}[3x + 5x] = 8\)

General Formulas and Concepts:

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           \(\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)\)

Basic Power Rule:

Step-by-step explanation:

Step 1: Define

Identify

\(\displaystyle y = 3x + 5x\)

Step 2: Differentiate

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Answer:

The point estimate for the true difference between the population means is 0.13.

The 90% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -0.35 years and 0.61 years.

Step-by-step explanation:

To solve this question, before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \(\mu\) and standard deviation \(\sigma\), the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \(\mu\) and standard deviation \(s = \frac{\sigma}{\sqrt{n}}\).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When we subtract two normal variables, the mean is the subtraction of the means while the standard deviation is the square root of the sum of the variances.

A sample of 263 cars owned by students had an average age of 7.25 years. The population standard deviation for cars owned by students is 3.77 years.

This means that:

\(\mu_s = 7.25, \sigma_s = 3.77, n = 263, s_s = \frac{3.77}{\sqrt{263}} = 0.2325\)

A sample of 291 cars owned by faculty had an average age of 7.12 years. The population standard deviation for cars owned by faculty is 2.99 years.

This means that:

\(\mu_f = 7.12, \sigma_f = 2.99, n = 291, s_f = \frac{2.99}{\sqrt{291}} = 0.1753\)

Difference between the true mean ages for cars owned by students and faculty.

Distribution s - f. So

\(\mu = \mu_s - \mu_f = 7.25 - 7.12 = 0.13\)

This is also the point estimate for the true difference between the population means.

\(s = \sqrt{s_s^2+s_f^2} = \sqrt{0.2325^2+0.1753^2} = 0.2912\)

90% confidence interval for the difference:

We have that to find our \(\alpha\) level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\(\alpha = \frac{1 - 0.9}{2} = 0.05\)

Now, we have to find z in the Ztable as such z has a pvalue of \(1 - \alpha\).

That is z with a pvalue of \(1 - 0.05 = 0.95\), so Z = 1.645.

Now, find the margin of error M as such

\(M = zs = 1.645*0.2912 = 0.48\)

The lower end of the interval is the sample mean subtracted by M. So it is 0.13 - 0.48 = -0.35 years

The upper end of the interval is the sample mean added to M. So it is 0.13 + 0.48 = 0.61 years.

The 90% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -0.35 years and 0.61 years.

a. The proportion or percentage of children with an MDI score of at least 120 is 10.56%.

b. The proportion or percentage of children with an MDI score of at least 80 is also 10.56%.

c. The MDI score that corresponds to the 99th percentile is approximately 137.

d. The MDI score such that only 1% of the population has an MDI score below it is approximately 63.

e. The lower quartile (Q1) of MDI scores is 89.28, and the upper quartile (Q3) is 110.72. These scores represent the range below which 25% of the population falls and the range above which 25% of the population falls, respectively.

f. The interquartile range (IQR) is 21.44.

g. The upper limit for potential outliers is 142.88, and the lower limit is 57.12. Any MDI scores outside this range would be considered potential outliers.

a. To find the proportion of children with an MDI score of at least 120, we need to calculate the z-score and use the standard normal distribution table. The z-score for an MDI score of 120 is (120-100)/16 = 1.25. Using the standard normal distribution table, the proportion of children with an MDI score of at least 120 is 0.1056 or 10.56%.

b. Similarly, to find the proportion of children with an MDI score of at least 80, we need to calculate the z-score for an MDI score of 80, which is (80-100)/16 = -1.25. Using the standard normal distribution table, the proportion of children with an MDI score of at least 80 is 0.1056 or 10.56%.

c. To find the MDI score that corresponds to the 99th percentile, we need to find the z-score that corresponds to the 99th percentile using the standard normal distribution table. The z-score is approximately 2.33. Using the formula z = (x - μ)/σ, we can solve for x to find the MDI score that corresponds to the 99th percentile: x = zσ + μ = 2.33 * 16 + 100 = 136.88. Therefore, the MDI score that corresponds to the 99th percentile is approximately 137.

d. To find the MDI score such that only 1% of the population has an MDI score below it, we need to find the z-score that corresponds to the 1st percentile using the standard normal distribution table. The z-score is approximately -2.33. Using the formula z = (x - μ)/σ, we can solve for x to find the MDI score that corresponds to the 1st percentile: x = zσ + μ = -2.33 * 16 + 100 = 63.12.

Therefore, the MDI score such that only 1% of the population has an MDI score below it is approximately 63.

e. The lower quartile (Q1) represents the 25th percentile, and the upper quartile (Q3) represents the 75th percentile. Using the standard normal distribution table, the z-score for the 25th percentile is approximately -0.67, and the z-score for the 75th percentile is approximately 0.67. Using the formula z = (x - μ)/σ, we can solve for x to find the MDI scores that correspond to Q1 and Q3: Q1 = -0.67 * 16 + 100 = 89.28, and Q3 = 0.67 * 16 + 100 = 110.72.

Therefore, the lower quartile of the MDI scores represents the scores below which 25% of the population falls, and the upper quartile represents the scores above which 25% of the population falls.

f. The interquartile range (IQR) is the difference between Q3 and Q1. Therefore, IQR = Q3 - Q1 = 110.72 - 89.28 = 21.44.

g. To find the upper and lower limits for potential outliers, we need to use the formula:

Upper limit = Q3 + 1.5IQR

Lower limit = Q1 - 1.5IQR

Substituting the values we have found, we get:

Upper limit = 110.72 + 1.521.44 = 142.88

Lower limit = 89.28 - 1.521.

The upper limit for potential outliers is 142.88, and the lower limit is 57.12. Any MDI scores outside this range would be considered potential outliers.

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Answer:

The boys' comic books are on the teacher's desk

comically

Step-by-step explanation:

Answer:

it is one to the right and 5 up compared to its parent function.

Step-by-step explanation:

the (x-1)3 means that the graph is shifted one to the right and the + 5 means that you have to shift the garph 5 up also.

The exact value of tan2θ in simplest radical form is  -21/√7.

The value of tan2θ at the given quadrant is calculated as follows;

From Pythagorean identity, we know that;

sin² θ + cos² θ = 1

cos² θ is calculated as follows;

(3/4)² + cos² θ = 1

9/16 + cos² θ = 1

cos² θ = 1 - 9/16

cos² θ = 7/16

cos θ = √(7/16)

The value of tan θ is calculated as follows;

tan θ = sin θ / cos θ = 3/4 x 4/√7 = 3/√7

Now, we will find tan 2θ;

tan 2θ = 2tan θ / (1 - tan² θ)

tan 2θ = 2(3/√7) / (1 - (3/√7)²)

tan 2θ = (6/√7) / (-2/7)

tan 2θ = -21/√7

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The completed ordered pairs are: (6, 29), (-1, 34), and (1, -6).

An ordered pair is of the form (x, y), which is defined by a relation y = f(x).

In the question, we are asked to determine the y-value of each ordered pair based on the given x-value, by the relation y = 7x - 13.

1. (6, _):

x-value = 6.

Substituting x = 6, in the relation y = 7x - 13, we get:

y = 7(6) - 13,

or, y = 42 - 13,

or, y = 29.

Thus, the complete ordered pair is (6, 29).

2. (-3, _):

x-value = -3.

Substituting x = -3, in the relation y = 7x - 13, we get:

y = 7(-3) - 13,

or, y = -21 - 13,

or, y = -34.

Thus, the complete ordered pair is (-1, -34).

3. (7, _):

x-value = 1.

Substituting x = 1, in the relation y = 7x - 13, we get:

y = 7(1) - 13,

or, y = 7 - 13,

or, y = -6.

Thus, the complete ordered pair is (1, -6).

Thus, the completed ordered pairs are: (6, 29), (-1, 34), and (1, -6).

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Answer:

m=54/5=104/5=10.8

Step-by-step explanation:

3.6=m×(1.2/3.6)

3.6=m×(12/36)

3.6=m×(1/3)=3.6

m=3.6×3

m=10.8

Done!

Answer:

Step-by-step explanation:

Set up an equation:

7x + 10x + 13x = 100000 and solve for x:

30x = 100000 so

x = 3333.33

Anna gets 7(3333.33) = 23333.31

Louise gets 10(3333.33) = 33333.33

Lacey gets 13(3333.33) = 43333.29

Answer:

1/2

Step-by-step explanation:

Combine the fractions by finding a common denominator.

2/2-1/2=1/2

1/2+1/2=2/2

Answer:

Extract form- 1/2

Decimals- 0.5

Step-by-step explanation:

Answer:

ok so x is how many she can buy so

4x≤25

x=6.26

so she can buy 6

Hope This Helps!!!

Answer:

1-x^2/x^3 - 1 = 1/x

Step-by-step explanation:

First rewrite x^3 as x^2 * x cancel x^2  in both numerator and denominator.

   Write below

1 - 1/x - 1

Subtract

   1 - 1 = 0

now simplify

1 -1 /x - 1 = 1/x

1/x  = Answer

Can you Understand

Answer:

75.36

Step-by-step explanation:

bc it sinple math if you think about it

Answer:

0.736 is the decimal form

Step-by-step explanation:

Answer:

the answer should be 0.736

hope that helped!

The Number of Soccer Balls is 16 and the Number of T-Shirts is 68

What is Linear Equation in Two Variables?

A linear equation in two variables is one that is stated in the form         ax + by + c = 0, where a, b, and c are real integers and the coefficients of x and y, i.e. a and b, are not equal to zero.

Solution:

Let,

Number of Soccer Balls = x

Number of T-Shirts = y

Equation 1:

x + y = 84 -----------(i)

Equation 2:

25x + 9.5y = 1046 ----------(ii)

Multiplying Equation 1 by 25

25x + 25y = 2100 ---------(iii)

Substracting Equation2 from Equation 1

15.5y = 1054

y = 68

So, x = 16

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The linear approximation used by the first step of Newton's method can be expressed as: f(x) ≈ f(a) + f'(a) * (x - a)

The linear approximation used by the first step of Newton's method is given by the equation:

f(x) ≈ f(a) + f'(a) * (x - a)

In other words, the linear approximation is produced by adding the derivative of the function at the initial guess and the difference between the updated estimate x and the starting guess a to the value of the function at the starting guess.

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The Intel pays $7,760 to Bally Manufacturing on August 14.

The first step in calculating what Intel Corporation pays Bally on August 14 is to determine the net price of the machinery after the trade discount and the return of $100 due to defects.

The trade discount of 40% is calculated as follows:

Discount = List price × Discount rate

Discount = $13,100 × 0.40 = $5,240

So the net price of the machinery after the trade discount is:

Net price = List price - Discount

Net price = $13,100 - $5,240 = $7,860

After Intel returns $100 of machinery, the cost of the machinery is further reduced to:

Net price after return = Net price - Return

Net price after return = $7,860 - $100 = $7,760

Since the payment terms are 3/10 EOM (end of month), Intel receives a discount of 3% if payment is made within 10 days. The 10-day period begins on August 1 and ends on August 10 (the payment due date). Since Intel pays the bill on August 14, payment is late and the 3% discount does not apply.

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Answer:

The probability that on a particular week, the hospital will receive on high BP pregnant woman is 0.0068

Step-by-step explanation:

We use the Exponential distribution,

Since we are given that on average, 5 pregnant women with high blood pressure come per week,

So, average = m = 5

Now, on average, 5 people come every week, so,

5 women per week,

so, we get 1 woman per (1/5)th week,

Hence, the mean is m = 1/5 for a woman arriving

and λ = 1/m = 5 = λ

we have to find the probability that it takes higher than a week for a high BP pregnant woman to arrive, i.e,

P(X>1) i.e. the probability that it takes more than a week for a high BP pregnant woman to show up,

Now,

P(X>1) = 1 - P(X<1),

Now, the probability density function is,

\(f(x) = \lambda e^{-\lambda x}\)

And the cumulative distribution function (CDF) is,

\(CDF = 1 - e^{-\lambda x}\)

Now, CDF gives the probability of an event occuring within a given time,

so, for 1 week, we have x = 1, and λ = 5, which gives,

P(X<1) = CDF,

so,

\(P(X < 1)=CDF = 1 - e^{-\lambda x}\\P(X < 1)=1-e^{-5(1)}\\P(X < 1)=1-e^{-5}\\P(X < 1) = 1 - 6.738*10^{-3}\\P(X < 1) = 0.9932\\And,\\P(X > 1) = 1 - 0.9932\\P(X > 1) = 6.8*10^{-3}\\P(X > 1) = 0.0068\)

So, the probability that on a particular week, the hospital will receive on high BP pregnant woman is 0.0068

Answer:

\(f(x) = - 2x - 1 \\ h(x) = - x + 5 \\ \\ The \: \: value \: \: of \: \: \: f(6) = ( - 2) \times 6 - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 13 \\ \\ as \: we \: know ,\: \\➪(hf)(x) = h(f(x)) \\ so \:➪( hf)(6) = h(f(6)) \\ \\ put \: the \: value \: \: of \: f(6) \: we \: \: get \\ \\➪ (hf)(6) \: = h( - 13) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - ( - 13) + 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 13 + 5 \\ \: \: \: \: \: \: \: = 18\)

Answer:

1) \(f'(t)=4t,\ g'(t)=3t^2+4\)

2) \(p(t) =2t^5+8t^3\)

\(p'(t)=10t^4+24t^2\)

3) False

4)\(q(t) =\dfrac{1}{2}t+2t^{-1}\)

\(q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}\)

5) False

Step-by-step explanation:

Given that:

\(f(t) = 2t^2\) and \(g(t) = t^3 + 4t\)

Formula:

\(1. \dfrac{d}{dx}x^n=nx^{n-1}\)

\(2. \dfrac{d}{dx}C.f(x)=C.f'(x)\ \{\text{C is a constant}\}\)

1) Using above formula:

\(f'(t)=2\times 2 t^{2-1}=4t\)

\(g'(t)=3t^{3-1}+4\times 1 t^{1-1}=3t^2+4\)

2) \(p(t) =2t^2(t^3+4t)\)

Rewriting the formula by distributing the \(2t^2\) term:

\(p(t) =2t^2.t^3+2t^2.4t=2t^5+8t^3\)

\(p'(t) = 10t^4+24t^2\)

3) By using answers of part (1):

\(f'(t).g'(t)=12t^3+16t\)

\(p'(t) = 10t^4+24t^2\)

Therefore it is False that \(p'(t) = f'(t).g'(t)\)

4) \(q(t)=\dfrac{t^3+4t}{2t^2}\)

Writing by distributing:

\(q(t)=\dfrac{t^3}{2t^2}+\dfrac{4t}{2t^2}\\\Rightarrow q(t) =\dfrac{t}{2}+\dfrac{2}{t}\\\Rightarrow q(t) =\dfrac{1}{2}t+2t^{-1}\)

Using the formula:

\(q'(t)=\dfrac{1}{2}t^{1-1}+2\dfrac{-1}{t^2}\\\Rightarrow q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}\)

(5)By using answers in part (1):

\(\dfrac{g'(t)}{f'(t)}=\dfrac{3t^2+4}{4t}=\dfrac{3}{4}t+\dfrac{1}t\)

\(q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}\)

Therefore, it is False that:

\(q'(t)=\dfrac{g'(t)}{f'(t)}\)

Step-by-step explanation:

Using the dividend discount model and the information given:

D1 = $2.00

beta = 0.9

risk-free rate = 4.2%

market risk premium = 4.5%

current stock price = $28.00

First, we need to find the required rate of return for the stock using the capital asset pricing model (CAPM):

r = r_rf + beta * (r_m - r_rf)

r = 0.042 + 0.9 * 0.045

r = 0.0831 or 8.31%

Next, we can use the dividend discount model to find the expected stock price at the end of 3 years:

D1 = D0 * (1 + g)

D0 = D1 / (1 + g)

P0 = D1 / (r - g)

P3 = D3 / (r - g)

P0 = current stock price = $28.00

D0 = D1 / (1 + g) = $2.00 / (1 + g)

D3 = D0 * (1 + g)^3

Substituting these values and solving for P3:

$28.00 = $2.00 / (1 + g) / (0.0831 - g)

$28.00 * (0.0831 - g) = $2.00 / (1 + g)

0.002343 - 0.028g + 0.0831g - 0.0831g^2 = 0.002343 + 0.002g

0.0831g^2 - 0.026g - 0.002343 = 0

g = 0.069 or 6.9%

Now we can use the dividend discount model again to find the expected stock price at the end of 3 years:

P3 = $2.00 * (1 + 0.069) ^ 3 / (0.0831 - 0.069)

P3 = $38.81

Therefore, the market believes the stock price at the end of 3 years will be $38.81

The sums are  38,32,79,40

According to the BODMAS rule, the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition, and at the end Subtraction. Solving any expression is considered correct only if the BODMAS rule or the PEMDAS rule is followed to solve it.

Given here: The addition of two numbers

When adding, always line up the addends, the two numbers being combined, one on top of each other according to their place values. Add the numbers in the ones column first, then the tens column, and finally the hundreds column, to get the sum, or the total.

1) 24+14=38

2) 7+25=32

3)76+3=79

4)26+14=40

Thus the sums of the respective equations gives 38,32,79,40 respectively

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Answer:

The first one means everything in the bracket is squared even the sign while the second one only the number is squared.

Step-by-step explanation:

Therefore answer for (-5)^(2) is 25

While for -5^2 is -25

Hope that helps

Answer:

838.89, simplified 840

Step-by-step explanation:

The roots of the quadratic function f(x) = -3x² - 11x - 5 will be negative 3.14 and negative 0.53.

Let the equation be ax² + bx + c = 0.

Then the roots of the equation will be

\(\rm x = \dfrac{-b \pm \sqrt{b^2 - 4 a c }}{2a}\)

The quadratic function is given below.

f(x) = -3x² - 11x - 5

Then the roots of the quadratic function will be given as,

x = [-(-11) ± √{(-11)² - 4 × (-3) × (-5)}] / [2 × (-3)]

Simplify the equation, then the roots of the equation are calculated as,

x = [-(-11) ± √{(-11)² - 4 × (-3) × (-5)}] / [2 × (-3)]

x = [ 11 ± √{121 - 60}] / [-6]

x = - (11 ± 7.81) / 6

x = - (11 + 7.81) / 6, - (11 - 7.81) / 6

x = - 3.14, -0.53

The solution of the quadratic function will be negative 3.14 and negative 0.53.

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Answer:

1) 2/9

2)3/9

Step-by-step explanation:

sorry,thats what i know so far

Answer:

\(m_{perpendicular}\) = 6

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

given

x + 6y = - 24 ( subtract x from both sides )

6y = - x - 24 ( divide through by 6 )

y = - \(\frac{1}{6}\) x - 4 ← in slope- intercept form

with slope m = - \(\frac{1}{6}\)

given a line with slope m then the slope of a line perpendicular to it is

\(m_{perpendicular}\) = - \(\frac{1}{m}\) = - \(\frac{1}{-\frac{1}{6} }\) = 6