An online polling site posed this question: "How much stock do you put in long-range weather forecasts?" Among its Web site users, 38, 528 chose to respond Complete parts (a) through (c) below.
a. Among the responses received, 3% answered with "a lot". What is the actual number of responses consisting of "a lot"?
b. Among the responses received, 18, 566 consisted of "very little or none". What percentage of responses consisted of "very little or none"?
c. Because the sample size of 38, 528 is so large, can we conclude that about 3% of the general population puts "a lot" of stock in long-range weather forecasts? Why or why not?
A. No, because the sample is a voluntary response sample, so the sample is not likely to be representative of the population.
B. Yes, because the sample is so large, the margin of error is negligible.
C. No, because even though the sample size is so large, there is still a margin of error.
D. Yes, because the sample size is large enough so that the sample is representative of the population.

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Answer:

Below

Step-by-step explanation:

● 8 ÷ (3-x)

Dividing by 3-x is like multiplying by 1/(3-x)

● 8 × (1/3-x)

● 8 /(3-x)

Answer:

X is uniformly distributed.

Step-by-step explanation:

Uniform Distribution:

This is the type of distribution where all outcome of a certain event have equal likeliness of occurrence.

Example of Uniform Distribution is - tossing a coin. The probability of getting a head is the same as the probability of getting a tail. The have equal likeliness of occurrence.

Answer:

1. cups of coffee sold

2.Y = 605.7 - 5.943x

3. -0.952

4. 70.84

Step-by-step explanation:

1. the dependent variable in this question is the cups of coffee sold

2. least square estimation line

Y = a+bx

we have y as the cups of coffee sold

x as temperature.

first we will have to solve for a and then b

∑X = 450

∑Y = 960

∑XY = 61600

∑X² = 35500

∑Y² = 221800

a = ∑y∑x²-∑x∑xy/n∑x²-(∑x)²

a = 960 * 35500-450*61600/6*35500-450²

a = 6360000/10500

= 605.7

b = n∑xy - ∑x∑y/n∑x²-(∑x)²

= 6*61600 - 450*960/6*35500 - 450²

= -5.943

the regression line

Y = a + bx

Y = 605.7 - 5.943x

3. we are to find correlation coefficient

r = n∑xy - ∑x∑y multiplied by√(n∑x²-(∑x)² * (n∑y² - (∑y)²)

= 6*61600 -960*450/√(6*35500 - 450²)*(6*221800 - 960²)

=-62400/√4296600000

= -62400/65548.5

= -0.952

4. we have to predict sales of a 90 degree day fro the regression line

Y = 605.7 - 5.943x

y = 605.7 - 5.943(90)

y = 605.7 - 534.87

= 70.84

Answer:

6.25

Step-by-step explanation:

2.5 *2.5=6.25

Answer:

6.25cm^2.

Step-by-step explanation:

To find the area of a square, you multiply the two sides, 2.5✖️2.5.

This gives the area of 6.25cm^2.

Hope this helped!

Have a nice day:)

Answer:

less than

Step-by-step explanation:

If the normality requirement is not satisfied​ (that is, ​np(1​ - p) is not at least​ 10), then a​ 95% confidence interval about the population proportion will include the population proportion in​ _less than__ 95% of the intervals.

The confidence interval consist of all reasonable values of a population mean. These are value for which the null hypothesis will not be rejected.

So, let assume that If the 95%  confidence interval contains the value for the hypothesized mean, then the sample mean  is reasonably close to the hypothesized mean. The effect of this is that the p- value is going to be greater than 0.05, so we fail to reject the null hypothesis.

On the other hand,

If the 95%  confidence interval do not contains the value for the hypothesized mean, then the sample mean  is far away from the hypothesized mean. The effect of this is that the p- value is going to be lesser than 0.05, so we reject the null hypothesis.

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

Answer:

this? hope it helps ........

Answer:

The answer is area=32pi-64 and the perimeter is 8pi

Step-by-step explanation:

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First we plot these point on a graph as given in attachment.

From the attachment we can observe that AD || BC || x-axis .

also, AB ||CD, that will make ABCD a parallelogram ,  but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".

Mid point of AC= [tex](\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Mid point of BD= [tex](\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Thus, Mid point of AC=Mid point of BD

i.e. diagonals bisect each other.

That means ABCD is a parallelogram.

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First, we plot these points on a graph as given in the attachment. From the attachment, we can observe that AD || BC || x-axis. Also, AB ||CD, which will make ABCD a parallelogram, but to confirm, we check the parallelogram property "diagonals bisect each other," i.e., "Midpoint of both diagonals is equal."

The midpoint of AC=. The midpoint of BD=. Thus, the Midpoint of AC=Mid point of BD diagonals bisects each other. That means ABCD is a parallelogram.

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Answer:

16.50

Step-by-step explanation:

Constant of proportionality = no of gallons of water per 1 minute.

In the first row, we have 16.50 gallons of water per 1 minute.

In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons

In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.

We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.

Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]

This means the constant of proportionality, 16.50, is same for all rows.

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Step-by-step explanation:

(f+g)(x) = f(x) + g(x)

= x/2-3 + 4x²+x+4

= ..........

[tex](f\circ g)(2)=\dfrac{1}{4\cdot2+9}=\dfrac{1}{17}\\\\(f+g)(2)=\dfrac{1}{2}+4\cdot2+9=\dfrac{1}{2}+17=\dfrac{1}{2}+\dfrac{34}{2}=\dfrac{35}{2}[/tex]

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Answer:

n + 1

Step-by-step explanation:

Starting with the integer 'n,' we represent the "next integer" by n + 1.

Answer:

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Step-by-step explanation:

Given

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100})[/tex]

Required

Simplify

For clarity, group the expression in threes

[tex]((\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the first group [Divide 8 by 4]

[tex]((\frac{3}{1})*(\frac{2}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 9 by 3]

[tex]((\frac{1}{1})*(\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex]((\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 15 by 3]

[tex]((\frac{2}{1})*(\frac{5}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 16 by 2]

[tex]((\frac{1}{1})*(\frac{5}{8}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the second group [Divide 35 and 25 by 5]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{7}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 49 by 7]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{1}{3})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 24 by 3]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{1}{1})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Merge the first and second group

[tex]((\frac{5}{8})*(\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](1*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the last group [Divide 99 by 9]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{9})*(\frac{11}{100}))[/tex]

[Divide 63 by 9]

[tex](\frac{4}{7})*((\frac{7}{64})*(\frac{80}{1})*(\frac{11}{100}))[/tex]

[Divide 64 and 80 by 8]

[tex](\frac{4}{7})*((\frac{7}{8})*(\frac{10}{1})*(\frac{11}{100}))[/tex]

[Divide 10 and 4 by 2]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{5}{1})*(\frac{11}{100}))[/tex]

[Divide 100 by 5]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{1}{1})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*(\frac{7}{4})*(\frac{11}{20})[/tex]

[tex]1*(\frac{11}{20})[/tex]

[tex]\frac{11}{20}[/tex]

Hence;

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Answer:

The upper bound of the confidence interval is 0.34

Step-by-step explanation:

Here in this question, we want to calculate the upper bound of the confidence interval.

We start by calculating the margin of error.

Mathematically, the margin of error = 0.29 -0.24 = 0.05

So to get the upper bound of the confidence interval, we simply add this margin of error to the mean

That would be 0.05 + 0.29 = 0.34

Answer:

30 cm

Step-by-step explanation:

let x be the lenght of the two sides of equal lenghts, so the other is x+3

and the perimeter is x+x +x +3

P=3x+3

P=3(x+1)

93=3(x+1)

31=x+1

x=30

so the shorter sides are of 30 centimeters and the longest is 33

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Answer:

A) C1 = 0.00187 m = 0.187 cm,  C2 = 0.0062 m = 0.62 cm

B)  A sample of how the graph looks like is attached below ( periodic sine wave )

C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum

Step-by-step explanation:

Given data :

mass = 5kg

length of spring = 10 cm = 0.1 m

f(t) = 10sin(t) N

viscous force = 2 N

speed of mass = 4 cm/s = 0.04 m/s

initial velocity = 3 cm/s = 0.03 m/s

Formulating initial value problem

y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m

spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m

f(t) = 10sin(t/2) N

using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion

the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)

A) finding the solution of the initial value

attached below is the solution and

B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like

C attached below

Answer:

40 minutes

Step-by-step explanation:

If it takes her 8 minutes to mow 1/6 of it, we can find the total amount of time it  will take by multiplying 8 by 6, since 1/6 times 6 is 1 (1 represents the whole lawn mowed)

8(6) = 48

The question asks for how many more minutes it will take, so subtract 48 by 8.

48 - 8 = 40

= 40 minutes

Answer:

40 minutes

Step-by-step explanation:

We can use ratios to solve

8 minutes          x minutes

------------------- = ----------------

1/6 yard                 1 yard

Using cross products

8 * 1 = 1/6 x

Multiply each side by 6

8*6 = 1/6 * x * 6

48 = x

48 minutes total

She has already done 8 minutes

48-8 = 40 minutes

Answer:

1/3(a+b)

hope it helps :>

Answer:

The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Step-by-step explanation:

The complete question is:

Salaries of 42 college graduates who took a statistics course in college have a​ mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard​ deviation, σ of ​$10​,016 construct a ​99% confidence interval for estimating the population mean μ.

Solution:

The (1 - α)% confidence interval for estimating the population mean μ is:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

The critical value of z for 99% confidence interval is:

[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]

Compute the 99% confidence interval for estimating the population mean μ as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

     [tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]

Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Answer:

300.05 miles

Step-by-step explanation:

initial fee= $39.99

final bill = $ 100

cost =$ 0.20 per mile

remaining amount = $ 60.01

solution,

she drive = remaining amount / cost

=60.01/0.20

=300.05 miles

Answer:

500 miles

Step-by-step explanation:

Let us use cross multiplication to find the unknown amount.

Given:

1) Cost for 1 mile=$0.20

2)Cost for x miles=$100

Solution:

No of miles                             Cost

1) 1                                             $0.20

2)x                                             $100

By cross multiplying,

100 x 1= 0.20x

x=100/0.20

x=500 miles

Thank you!