In a factory there are 100 units of a certain product, 5 of which are defective. We pick three units from the 100 units at random. What is the probability that none of them are defective

Answers

Answer 1

Answer:

Probability of picking all three non-defective units

= 7372/8085  (or 0.911812 to six decimals)

Step-by-step explanation:

Let

D = event that the picked unit is defective

N = event that the picked unit is not defective

Pick are without replacement.

We need to calculate P(NNN) using the multiplication rule,

P(NNN)

= 97/100 * 96/99 * 95/98

=7372/8085

= 0.97*0.969697*0.9693878

= 0.911812

Answer 2

The probability that none of the picked products are defective is;

P(None picked is defective) = 0.856

We are told that 5 are defective out of 100.

This means the number of good products that are not defective are 95.

Probability of the first picked product not being defective is written as; P(First picked not defective) = 95/100

Since the good ones have been picked, there will be 99 left of which the good ones are now 94. Thus, probability of second one not being defective = 94/99

Since two good ones have been picked, there will be 98 left and 93 good ones left. Thus, probability of third one not being defective = 93/98

Finally, Probability of none of the three being defective is;

95/100 × 94/99 × 93/98 = 0.856

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Related Questions

A machine used to fill​ gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be​ reset? Explain your reasoning. ▼ Yes No ​, it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ​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.

Answers

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.

5x+4(-x-2)=-5x+2(x-1)+12

Answers

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

please help me in these question ????

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
(b) How many samples have 3 red pens and 1 black pen?
(c) How many samples of size 4 contain at least two red pens?
(d) How many samples of size 4 contain


If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal distribution.
1- What percentage of a cucumber give the crop amount between and 834 kg?
2- What the probability of cucumber give the crop exceed 900 kg ?

Answers

Answer:

Step-by-step explanation:

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.

(a) How many different samples of size 4 pens are possible?

12C4=12!/(4!*8!)=495

(b) How many samples have 3 red pens and 1 black pen?

5C3*7C1

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

=>5C3*7C1=10*7=70

(c) How many samples of size 4 contain at least two red pens?

(5C2*7C2)+(5C3*7C1)+(5C4*7C0)

5C2=5!/(2!*3!)=10

7C2=7!/(2!*5!)=21

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

5C4=5!/(4!*1!)=5

7C0=7!/(0!*7!)=1

=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285

(d) How many samples of size 4 contain at most one black pen?

(7C1*5C3)+(7C0*5C4)

7C1=7!/(1!*6!)=7

7C0=7!/(0!*7!)=1

5C3=5!/(3!*2!)=10

5C4=5!/(4!*1!)=5

=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75

The cost of a daily rental car is as follows: The initial fee is $39.99 for the car, and it costs $0.20 per mile. If Julie's final bill was $100.00 before taxes, how many miles did she drive?

Answers

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!

Which of the following represents "next integer after the integer n"? n + 1 n 2n

Answers

Answer:

n + 1

Step-by-step explanation:

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

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​ ________ 95% of the intervals. ​(This is a reading assessment question. Be certain of your answer because you only get one attempt on this​ question.)

Answers

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.

find the area of square whose side is 2.5 cm

Answers

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:)

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
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?

Answers

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.

The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:

Answers

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]

       

 

Salaries of 42 college graduates who took a statistics course in college have a​ mean, ​, of . Assuming a standard​ deviation, ​, of ​$​, construct a ​% confidence interval for estimating the population mean .

Answers

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).

a
A solid metal cone of base radius a cm and height 2a cm is melted and solid
spheres of radius are made without wastage. How many such spheres can be
made?​

Answers

volume of a cone

.

.

.

volume of sphere

.

.

number of spheres that can be made......

.

.

hence a hemisphere can be formed

Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Answers

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]

Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente

Answers

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

X = 1mL

A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10sin( t ) N(newtons) and moves in a medium that imparts a viscous force of 2 N
when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.
A)Find the solution of the initial value problem in the above problem.
B)Plot the graph of the steady state solution
C)If the given external force is replaced by a force of 2 cos(ωt) of frequency ω , find the value of ω for which the amplitude of the forced response is maximum.

Answers

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

The following shape is based only on squares, semicircles, and quarter circles. Find the area of the shaded part.

Answers

Answer:

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

Answer:

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

Step-by-step explanation:

Transform the given parametric equations into rectangular form. Then identify the conic.

Answers

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.

Time

(minutes)

Water

(gallons)

1

16.50

1.5

24.75

2

33

find the constant of proportionality for the second and third row

Answers

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.

The quotient of 8 and the difference of three and a number​.
Answer: 8÷(3-x)

Answers

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)

Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...

Answers

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

Let X denote the day she gets enrolled in her first class and let Y denote the day she gets enrolled in both the classes. What is the distribution of X

Answers

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.

88 feet/second = 60 miles/hour. How many feet per second is 1 mile/hour? (Hint: divide both sides of the equation
by the same amount.)
Round to the nearest thousandth.
One mile per hour is equivalent to
ao feet/second

Answers

Answer: 1ft/sec = 0.618 mi/hr

Explanation:

88 ft/sec = 60 mi/hr
88/88 ft/sec = 60/88 mi/hr (divide both sides by 88)
1 ft/sec = 60/88 mi/hr
1 ft/sec = 15/22 mi/hr
1 ft/sec = 0.681 mi/hr

Kenji earned the test scores below in English class.
79, 91, 93, 85, 86, and 88
What are the mean and median of his test scores?

Answers

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

Answers

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]

one third multiplied by the sum of a and b

Answers

Answer:

1/3(a+b)

hope it helps :>

a+b/3
This is the answer of ur question

logx-log(x-l)^2=2log(x-1)​

Answers

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.

Max believes that the sales of coffee at his coffee shop depend upon the weather. He has taken a sample of 5 days. Below you are given the results of the sample.
Cups of Coffee Sold Temperature
350 50
200 60
210 70
100 80
60 90
40 100
A. Which variable is the dependent variable?
B. Compute the least squares estimated line.
C. Compute the correlation coefficient between temperature and the sales of coffee.
D. Predict sales of a 90 degree day.

Answers

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

What is the most precise name for quadrilateral ABCD with vertices A(–5,2), B(–3, 5),C(4, 5),and D(2, 2)?

Answers

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.

Suppose that a sample mean is .29 with a lower bound of a confidence interval of .24. What is the upper bound of the confidence interval?

Answers

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

HELP ASAP ROCKY!!! will get branliest.​

Answers

Answer:

work pictured and shown

Answer:

Last one

Step-by-step explanation:

● [ ( 3^2 × 5^0) / 4 ]^2

5^0 is 1 since any number that has a null power is equal to 1.

●[ (3^2 ×1 ) / 4 ]^2

● (9/4)^2

● 81 / 16

The quotient of 3 and the cube
of y+2

Answers

Answer:

  [tex]\dfrac{3}{(y+2)^3}[/tex]

Step-by-step explanation:

Maybe you want this written using math symbols. It will be ...

  [tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]

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