The method of tree-ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

1,2851,1871,2221,1941,2681,3161,2751,3171,275

Required:
a. Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviations.
b. Find a 90% confidence interval for the mean of all tree-ring dates from this archaeological site. (Round your answers to the nearest whole number.)

9514 1404 393

Answer:

Step-by-step explanation:

1. Input 3 for x and do the arithmetic.

  z(x) = x+1

  z(3) = 3+1 = 4 . . . . . the output is 4

__

2. The expression for n(x) has x in the denominator. The expression will be undefined when the denominator is zero, so x=0 cannot be used.

Answer:

2 pounds

Step-by-step explanation:

so if we say we decrease 20% each year and we're only going one year. All you have to do is 2.5 - 20%

there are 12 set of months in a year

Answer:

I think the right answer is: Acute

Answer:

[tex]x = 3[/tex]

Step-by-step explanation:

Given

[tex]\log6^{(4x-5)} =\log6^{(2x+1)}[/tex]

Required

Solve for x

We have:

[tex]\log6^{(4x-5)} =\log6^{(2x+1)}[/tex]

Remove log6 from both sides

[tex](4x-5) = (2x+1)[/tex]

Collect like terms

[tex]4x - 2x = 5 + 1[/tex]

[tex]2x = 6[/tex]

Divide by 2

[tex]x = 3[/tex]

Answer:

60, 37, 34, 28, 40

(D)

ED2021

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

[tex]\displaystyle y^2 = 1 + \sin x[/tex]

And we want to prove that:

[tex]\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1[/tex]

Find the first derivative by taking the derivative of both sides with respect to x:

[tex]\displaystyle 2y \frac{dy}{dx} = \cos x[/tex]

Divide both sides by 2y:

Find the second derivative using the quotient rule:

[tex]\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\ &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}[/tex]

Substitute:

[tex]\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right) + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1[/tex]

Simplify:

[tex]\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1[/tex]

Combine fractions:

[tex]\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1[/tex]

Simplify:

[tex]\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1[/tex]

Cancel:

[tex]\displaystyle -\sin x + y^2 = 1[/tex]

Substitute:

[tex]-\sin x + \left( 1 + \sin x\right) =1[/tex]

Simplify. Hence:

[tex]1\stackrel{\checkmark}{=}1[/tex]

Q.E.D.

D

A

B

C

for more explanation please don't hesitate to just respond

Answer:

[tex]{ \tt{hypotenuse = { \boxed{5}}}} \\ { \tt{opposite = { \boxed{3}}}} \\ { \tt{adjacent = { \boxed{4}}}} \\ \\ { \tt{ \sin \angle R = \frac{{ \boxed{3}}}{{ \boxed{5}}} }} \\ \\ { \tt{ \cos \angle R = \frac{{ \boxed{4}}}{{ \boxed{5}}} }} \\ \\ { \tt{ \tan \angle R = \frac{ \boxed{3}}{{ \boxed{4}}} }}[/tex]

Answer:

260% is the correct answer

Step-by-step explanation:

i hope i helped

Answer:

60

Step-by-step explanation:

10 x 6 = 60

Hope this helped! :)

Two workers finished a job in 7.5 days.

How long would it take each worker to do the job by himself if one of the workers needs 8 more days to finish the job than the other worker?

let t = time required by one worker to complete the job alone

then

(t+8) = time required by the other worker (shirker)

let the completed job = 1

A typical shared work equation

7.5%2Ft + 7.5%2F%28%28t%2B8%29%29 = 1

multiply by t(t+8), cancel the denominators, and you have

7.5(t+8) + 7.5t = t(t+8)

7.5t + 60 + 7.5t = t^2 + 8t

15t + 60 = t^2 + 8t

form a quadratic equation on the right

0 = t^2 + 8t - 15t - 60

t^2 - 7t - 60 = 0

Factor easily to

(t-12) (t+5) = 0

the positive solution is all we want here

t = 12 days, the first guy working alone

then

the shirker would struggle thru the job in 20 days.

Answer:7 + 17 = 24÷2 (since there are 2 workers) =12. Also, ½(7) + ½17 = 3.5 + 8.5 = 12. So, we know that the faster worker will take 7 days and the slower worker will take 17 days. Hope this helps! jul15

Step-by-step explanation:

Step-by-step explanation:

11mg_6.5mg=4.5mg

in 24mins 4.5mgwas gone.

6.5mg_2mg=4.5mg

you have to lose 4.5mg more

answer 24+ 24

48 minutes

Answer:

Blair & Rosen (B&R) Plc.

Recommendation for moderate investor:

Internet fund = 96/240 * $500,000 = $200,000

Blue-chip fund = 144/240 * $500,000 = $300,000

Annual return for the portfolio:

Internet fund = $200,000 * 12% =  $24,000

Blue-chip fund = $300,000 * 9% = $27,000

Total portfolio returns =                  $51,000

Annual returns of portfolio = $51,000/$500,000 * 100 = 10.2%

Recommendation for aggressive investor:

Internet fund = 192/320 * $500,000 = $300,000

Blue-chip fund = 128/320 * $500,000 = $200,000

Step-by-step explanation:

a) Data and Calculations:

Maximum investible savings = $500,000

Projected annual return of the internet fund = 12%

Projected annual return of the blue-chip fund = 9%

Maximum determined amount to invest in the internet fund = $350,000

Risk rating for the internet fund = 6/1,000

Risk rating for the blue-chip fund = 4/1,000

Maximum risk rating for a moderate investor = 240

Maximum risk rating for an aggressive investor = 320

Recommendation for moderate investor:

Internet fund = 96/240 * $500,000 = $200,000

Blue-chip fund = 144/240 * $500,000 = $300,000

Annual return for the portfolio:

Internet fund = $200,000 * 12% = $24,000

Blue-chip fund = $300,000 * 9% = $27,000

Total returns = $51,000

Annual returns of portfolio = $51,000/$500,000 * 100 = 10.2%

Recommendation for aggressive investor:

Internet fund = 192/320 * $500,000 = $300,000

Blue-chip fund = 128/320 * $500,000 = $200,000

We can split 200 into 100 x 2.

100 is a perfect square and its square root is 10. The 2 will remain under the radical.

ANSWER: 10 sqrt(2)

(Option 2)

Hope this helps!

Answer:

x = 10

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Equality Properties

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

14x - (-5x + 6 - 3x - 4) = 4(5x) + 10

Step 2: Solve for x

Answer: a = 25.67

Step-by-step explanation:

Var[2X + 3Y] = 2² Var[X] + 2 Cov[X, Y] + 3² Var[Y]

but since X and Y are given to be independent, the covariance term vanishes and you're left with

Var[2X + 3Y] = 4 Var[X] + 9 Var[Y]

X follows an exponential distribution with parameter λ = 1/6, so its mean is 1/λ = 6 and its variance is 1/λ² = 36.

Y is uniformly distributed over [a, b] = [4, 10], so its mean is (a + b)/2 = 7 and its variance is (b - a)²/12 = 3.

So you have

Var[2X + 3Y] = 4 × 36 + 9 × 3 = 171

Answer:

32

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

2 * 3 to the -2 power. 3x-3=-9

2x-9=-18

Answer:

The p-value of the test is 0.242 > 0.05, which means that this information does not indicate a difference between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

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]

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Women:

51 out of 222, so:

[tex]p_1 = \frac{51}{222} = 0.2297[/tex]

[tex]s_1 = \sqrt{\frac{0.2297*0.7703}{222}} = 0.0282[/tex]

Men:

49 out of 174, so:

[tex]p_2 = \frac{49}{174} = 0.2816[/tex]

[tex]s_2 = \sqrt{\frac{0.2816*0.7184}{174}} = 0.0341[/tex]

Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts?

Either way, so a two tailed test to see if the difference of proportions is different of 0.

At the null hypothesis, we test if it is not different of 0, so:

[tex]H_0: p_1 - p_2 = 0[/tex]

At the alternative hypothesis, we test if it is different of 0, so:

[tex]H_1: p_1 - p_2 \neq 0[/tex]

The test statistic is:

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

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that [tex]\mu = 0[/tex]

From the samples:

[tex]X = p_1 - p_2 = 0.2297 - 0.2816 = -0.0519[/tex]

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0282^2+0.0341^2} = 0.0442[/tex]

Value of the test statistic:

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

[tex]z = \frac{-0.0519 - 0}{0.0442}[/tex]

[tex]z = -1.17[/tex]

P-value of the test and decision:

The p-value of the test is the probability of the differences being of at least 0.0519, either way, which is P(|z| > 1.17), that is, 2 multiplied by the p-value of z = -1.17.

Looking at the z-table, z = -1.17 has a p-value of 0.121.

0.121*2 = 0.242

The p-value of the test is 0.242 > 0.05, which means that this information does not indicate a difference between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts.

y'' - 6y' + 9y = 0

If y = C₁ exp(3x) + C₂ x exp(3x), then

y' = 3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x))

y'' = 9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x))

Substituting these into the DE gives

(9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x)))

… … … - 6 (3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x)))

… … … + 9 (C₁ exp(3x) + C₂ x exp(3x))

= 9C₁ exp(3x) + 6C₂ exp(3x) + 9C₂ x exp(3x))

… … … - 18C₁ exp(3x) - 6C₂ (exp(3x) - 18x exp(3x))

… … … + 9C₁ exp(3x) + 9C₂ x exp(3x)

= 0

so the provided solution does satisfy the DE.

9514 1404 393

Answer:

  (-13, 10)

Step-by-step explanation:

If M is the midpoint of segment DE, then ...

  D = 2M -E

  D = 2(-3, 4) -(7, -2) = (2(-3)-7, 2(4)+2) = (-13, 10)

The other end point is (-13, 10).

9514 1404 393

Answer:

  split the number into equal pieces

Step-by-step explanation:

Assuming "splitting any number" means identifying parts that have the number as their sum, the maximum product of the parts will be found where the parts all have equal values.

We have to assume that the number being split is positive and all of the parts are positive.

If we divide number n into parts x and (n -x), their product is the quadratic function x(n -x). The graph of this function opens downward and has zeros at x=0 and x=n. The vertex (maximum product) is halfway between the zeros, at x = (0 + n)/2 = n/2.

Similarly, we can look at how to divide a (positive) number into 3 parts that have the largest product. Let's assume that one part is x. Then the other two parts will have a maximum product when they are equal. Their values will be (n-x)/2, and their product will be ((n -x)/2)^2. Then the product of the three numbers is ...

  p = x(x^2 -2nx +n^2)/4 = (x^3 -2nx^2 +xn^2)/4

This will be maximized where its derivative is zero:

  p' = (1/4)(3x^2 -4nx +n^2) = 0

  (3x -n)(x -n) = 0 . . . . . . . . . . . . . factor

  x = n/3  or  n

We know that x=n will give a minimum product (0), so the maximum product is obtained when x = n/3.

A similar development can prove by induction that the parts must all be equal.

Answer:hi

Step-by-step explanation:

Answer:

[tex]{ \tt{ = ( log_{x}16)( log_{2}x) }}[/tex]

Change base x to base 2:

[tex]{ \tt{ = (\frac{ log_{2}16}{ log_{2}x } )( log_{2}x)}} \\ \\ { \tt{ = log_{2}(16) }} \\ = { \tt{ log_{2}(2) }} {}^{4} \\ = { \tt{4 log_{2}(2) }} \\ = { \tt{4}}[/tex]

Answer:

B) All real numbers except 0 and integer multiples of 8π∕5

Step-by-step explanation:

Cotangent function:

The cotangent function is given by:

[tex]y = \cot{ax} = \frac{\cos{ax}}{\sin{ax}}[/tex]

Domain:

All real values except those at which:

[tex]\sin{ax} = 0[/tex]

The sine is 0 for 0 and all integer multiples of [tex]\frac{1}{a}[/tex]

In this question:

[tex]a = \frac{5}{8}[/tex], so the values outside the domain are 0 and the integer multiples of [tex]\frac{8}{5}[/tex]. Then the correct answer is given by option b.

Answer:

Positive numbers.

Step-by-step explanation:

Numbers after zero are positive numbers, which can be any number (whole or  decimal/fraction). But numbers before zero are negative numbers which can be also whole or decimal fraction.

Example for numbers to the right of 0: 7, 6.5, 8/10