f(x)=2x1 + 16x2 + 7x3 + 4x4 -> min

===============================================

Work Shown:

A = P*e^(r*t)

A = 47000*e^(0.0526*17)

A = 114,932.799077198

A = 114,932.80

Notes:

Mark's account balance after 17 years would be $114,932.8

[tex]A=Pe^{rt}[/tex]

where,

A = Accrued amount

P = Principal amount

r = interest rate as a decimal

R = interest rate as a percent

r = R/100

t = time in years

For given question,

P = $47000, t = 17 years

R = 5.26%

[tex]\Rightarrow r =\frac{5.26}{100}\\\\\Rightarrow r = 0.0526[/tex]

Using the Continuous Compounding Formula,

[tex]\Rightarrow A=Pe^{rt}\\\\\Rightarrow A=47000\times e^{0.0526\times 17}\\\\\Rightarrow A=114932.8[/tex]

Therefore, Mark's account balance after 17 years would be $114,932.8

Learn more about the Continuous Compounding here:

https://brainly.com/question/24246899

#SPJ2

Answer:

x = 2.7

Step-by-step explanation:

The given equation is :

[tex](3x)^2-10=56[/tex]

We need to solve it for x.

It can be rewrite as follows:

[tex]9x^2-10=56[/tex]

Adding 10 to both sides,

[tex]9x^2-10+10=56+10\\\\9x^2=66\\\\x=\sqrt{\dfrac{66}{9}}\\\\x=2.70[/tex]

So, the value of x is equal to 2.7.

Hi there!

[tex]\large\boxed{(x + 2)}[/tex]

x² + 3x + 2

We know that x + 1 is a factor, so:

We must find another number that adds up to 3 when added to 1 and multiplies into 2 with 1. We get:

x + 2

(x + 1)(x + 2)

Answer:

10^3000

Step-by-step explanation:

1000¹⁰⁰⁰

= (10³)¹⁰⁰⁰

= 10^(3×1000)

= 10³⁰⁰⁰ or 10^3000

Answered by GAUTHMATH

Answer:

the answer should be e^3

Step-by-step explanation:

i hope it helps you

Answer:

4,a

5.d

6.c

plz mark me as brainliest

Step-by-step explanation:

Answer:

1. A

2. C

3. A

Step-by-step explanation:

all the explanations are In the image above

9514 1404 393

Answer:

  $4000

Step-by-step explanation:

The problem tells you the relation between commission (c) and stock value (v) is ...

  c = 10 + 0.025v . . . . $10 + .025 of the value traded

We want to find the value (v) for the given commission (c=110). We can put these numbers into the formula and solve for v:

  110 = 10 + 0.025v

  100 = 0.025v . . . . . . . subtract 10

  100/0.025 = v = 4000 . . . . divide by the coefficient of v

The value of stock traded was $4000.

_____

Additional comment

As always, you start by reading and comprehending the problem. You look for what is being asked for, what is being given, and any information that relates one to the other.

Here, the value of stock traded is asked for, the amount of commission is given, and a description of the relation of one to the other is provided. Translate that description to an equation, fill in the given value, and solve for the unknown. (That's what we did above.)

__

You can also work this in your head. The commission is $10 more than some fraction of the amount traded. Since the commission is $110, only $100 of that is the fraction of the amount traded. With a little experience, you can recognize the fraction 0.025 as being 1/40. That means $100 is 1/40 of the value traded, or the value traded is 40 × $100 = $4000.

Answer:

a) 0.15 / 0.09

b) 0.15 / 1

c) 0.15 / 0.23

⭕ 17.3

#CARRYONLEARNING

[tex]{hope it helps}}[/tex]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

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

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]

Sample of 10:

This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]

[tex]Z = 1.28[/tex]

[tex]Z = 1.28[/tex] has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

[tex]\mu = 266, \sigma = 16[/tex]

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{260 -  266}{16}[/tex]

[tex]Z = -0.375[/tex]

[tex]Z = -0.375[/tex] has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 20[/tex], so:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]

[tex]Z = -1.68[/tex]

[tex]Z = -1.68[/tex] has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 50[/tex], so:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]

[tex]Z = -2.65[/tex]

[tex]Z = -2.65[/tex] has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = 2.42[/tex]

[tex]Z = 2.42[/tex] has a p-value of 0.9922.

X = 256

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = -2.42[/tex]

[tex]Z = -2.42[/tex] has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step 1: Find the circumference of the circle

Formula for circumference: C = 2 * pi * r

C = 2 * pi * 27

C = 54pi

Step 2: Find the length of the desired arc

We are only looking for 20 degrees out of 360 degrees, therefore we can multiply our circumference by 20/360.

20/360 * 54pi = 3pi units

Hope this helps!

Answer:

The equation is [tex]A(d) = 80 - 4d[/tex]

Step-by-step explanation:

Linear function:

A linear function for the amount of money in an account after t days is given by:

[tex]A(d) = A(0) - md[/tex]

In which A(0) is the initial value and m is the daily cost.

Your EZ Pass account begins with $80. It costs you $4/day.

This means that [tex]A(0) = 80, m = 4[/tex]

So

[tex]A(d) = A(0) - md[/tex]

[tex]A(d) = 80 - 4d[/tex]

Answer:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Required

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex]

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] becomes

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Multiply by 1

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}\cdot 1[/tex]

Express 1 as

[tex]\frac{y^2}{y^2} = 1[/tex]

So, the expression becomes:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} \cdot \frac{y^2}{y^2}[/tex]

Rewrite as:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} \cdot \frac{\sin^2y}{y^2}[/tex]

In limits:

[tex]\lim_{(x,y) \to (0,0)} \frac{\sin^2y}{y^2} \to 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} *1[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2}[/tex]

Convert to polar coordinates; such that:

[tex]x = r\cos\theta;\ \ y = r\sin\theta;[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{(r\cos\theta)^2 (r\sin\theta;)^2}{(r\cos\theta)^2+2(r\sin\theta;)^2}[/tex]

Expand

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2\cos^2\theta+2r^2\sin^2\theta}[/tex]

Factor out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2(\cos^2\theta+2\sin^2\theta)}[/tex]

Cancel out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

Express [tex]2\sin^2 \theta[/tex] as [tex]\sin^2\theta+\sin^2\theta[/tex]

So:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+\sin^2\theta+\sin^2\theta}[/tex]

In trigonometry:

[tex]\cos^2\theta + \sin^2\theta = 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

Evaluate the limits by substituting 0 for r

[tex]\frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0}{1+\sin^2\theta}[/tex]

Since the denominator is non-zero; Then, the expression becomes 0 i.e.

[tex]\frac{0}{1+\sin^2\theta} = 0[/tex]

So,

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Answer:

D. 750(0.75)ˣ

Step-by-step explanation:

Let the new brightness be L'. Since our initial brightness L₀ reduces by 25 %, we have that L' = L₀ - 25% of L₀

L' = L₀ - 0.25L₀

L' = 0.75L₀

Adding the second screen, the new intensity is L" = L' - 25 % of L'  

L" = L' - 0.25 L'

L" = 0.75L'.

Since L' = 0.75L₀,

L" = 0.75L' = 0.75(0.75L₀) = 0.75²L₀

Adding the third screen, the new intensity is L"' = L'' - 25 % of L''  

L'" = L" - 0.25 L"

L"' = 0.75L".

Since L" = 0.75L' = 0.75²L₀

L"' = 0.75L" = 0.75(0.75²L₀) = 0.75³L₀

So, we see a pattern here.

The intensity after x screens is L = (0.75)ˣL₀

Since L₀ = 750 lumens,

L = 750(0.75)ˣ

Answer:

[tex]x = 7[/tex]

Step-by-step explanation:

Given

[tex]AC = 3x +3[/tex]

[tex]AB = -1+2x\\[/tex]

[tex]BC =11[/tex]

Required

Find x

We have:

[tex]AC = AB + BC[/tex]

So, we have:

[tex]3x + 3 = -1 + 2x + 11[/tex]

Collect like terms

[tex]3x - 2x = -1 + 11 - 3[/tex]

[tex]x = 7[/tex]

Answer:

15

Step-by-step explanation:

3+7+5

PLEASE BRAINLIESTTTTTTTTT

Answer:

c. 5.5 years, 0.51 years

Step-by-step explanation:

Central Limit Theorem

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

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Mean of 5.5 years and a standard deviation of 2.1 years.

This means that, for the population, [tex]\mu = 5.5, \sigma = 2.1[/tex]

Random samples of size 17.

This means that [tex]n = 17[/tex]

Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution.

The mean is the same as the mean for the population, that is, 5.5 years.

The standard deviation is:

[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{17}} = 0.51[/tex]

This means that the correct answer is given by option c.

Answer:

[tex]Sales = 86.749[/tex]

Step-by-step explanation:

Given

[tex]Sales = 0.845*(competitors) + 5.79*(population) + 13.889[/tex]

[tex]Competitors = 4[/tex]

[tex]Population = 12000[/tex]

See comment for complete question

Required

The sales

We have:

[tex]Sales = 0.845*(competitors) + 5.79*(population) + 13.889[/tex]

Substitute values for competitors and population

[tex]Sales = 0.845*4 + 5.79*12 + 13.889[/tex]

[tex]Sales = 3.38 + 69.48 + 13.889[/tex]

[tex]Sales = 86.749[/tex]

Answer:

1.125 hours

Step-by-step explanation:

Given :

Ben's speed = 3 mi/hr

Time before Amanda starts = 1.5 hours

Amanda's speed = 7 mi/hr

Time before Amanda catches up with Ben

Recall :

Distance = speed * time

Distance already covered by Ben before Amanda starts :

(3 * 1.5) = 4.5

Hence, we can setup the equation :

Ben's distance = Amanda's distance

Let time taken = x

4.5 + 3x = 7x

4.5 = 7x - 3x

4.5 = 4x

x = 4.5 / 4

x = 1.125 hours

1.125 * 60 = 67. 5 minutes

Answer:

Following are the response to the given choice.

Step-by-step explanation:

Please find the complete question in the attached file.

Subjective opinion = Question of opinion.

Therefore this requires only just opinion and we don't have to do any actual calculations.

Does this seem like a great number of girls, a little number of girls, or a decent number of girls to you but if 1,300 babies were born, 5 of whom were females?

This is a small number of beautiful gals, in my honest opinion. We anticipate boys and girls to be produced about the very same frequency, thus I expect some half of them to be females if there are 1 300 newborns. You should have roughly 650 girls if 50% of the infants are girls, but now we only have five. That appears to me to be considerably low. which is your own opinion.

Answer:

hope this will help you a lot

Answer:

20 participants

Step-by-step explanation:

Given

[tex]Conditions = 2[/tex]

[tex]Scores = 20[/tex]

Type: Repeated design

Required

The number of participants (n)

The repeated measure design implies that the test was conducted repeatedly on the same sample size.

Since the score in each test is 20; then:

[tex]n = 20[/tex] --- the number of participants

If z ⁷ = 128i, then there are 7 complex numbers z that satisfy this equation.

[tex]z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}[/tex]

[tex]\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}[/tex]

(where n = 0, 1, 2, …, 6)

[tex]\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}[/tex]

[tex]\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)[/tex]

Answer:

C. 2 and 3 opens down

Step-by-step explanation:

see graph

Answer:

3x² + 13x + 4

Step-by-step explanation:

I did the steps in my book

Answer:

15

Hope this helped ^^

2[30          5[45

 3[15           3[9

  5[5             3[3

     [1               [1

    30=2*3*5*1

     45=5*3*3*1    HCF= 3*1=3

hope its helps you

keep smiling be happy stay safe

Answer:

where is the statement

Step-by-step explanation:

its incomplete po

Step-by-step explanation:

true