Solve each equation for x.
a(x-b)+bx+c=a+b

Answer:

2.45

Step-by-step explanation:

Given that :

α = 0.05

Larger sample variance= numerator, sample size = 9

Smaller sample variance = denominator, sample size = 21

Hence,

DFnumerator = n - 1 = 9 - 1 = 8

DFdenominator = n - 1 = 21 - 1 = 20

Critical value for upper tail test using the F distribution table at α = 0.05 ; DFnumerator on horizontal ; Df denominator as vertical ;

F critical = 2.447

F critical = 2.45

Answer:

d. The average amount of weight loss is greater than 2.5 pounds.

Step-by-step explanation:

After running a mile a day over a period of two weeks, the average amount of weight loss is 2.5 pounds.

At the null hypothesis, we test if this mean is of 2.5, that is:

[tex]H_0: \mu = 2.5[/tex]

A dietitian, who publishes health articles in a newspaper, states their new diet program helps with additional weight loss.

With the additional weight loss, the dietitian claims that the mean is more than the value presented at the null hypothesis, that is, more than 2.5, and thus, the correct answer is:

[tex]H_1: \mu > 2.5[/tex]

And thus, the correct option is given by option d.

Answer:

n= (z)22E2

n=10× 99%÷ 0.07

Answer:

10/16=5/8

6+6+4=16

The probability is 5/8

Answer:

Years = natural log (Total / Principal) / Rate

Years = natural log (1,000,000 / 2,500) / .02

Years = natural log (400) / .02

Years =  5.9914645471 / .02

It would take  299.573227355 Years

Source: http://www.1728.org/rate2.htm

Step-by-step explanation:

The identity as been verified/proved as:

[tex]1 - \tan\ a\ tan\ b = 1 - \tan\ a\ tan\ b[/tex]

Given that:

[tex]\frac{\cos(a + b)}{\cos\ a\cos b} = 1 - \tan\ a\ tan\ b[/tex]

Apply cosine identity to the numerator

[tex]\frac{\cos\ a\ cos\ b - \sin a\ sin\ b}{\cos\ a\cos b} = 1 - \tan\ a\ tan\ b[/tex]

Split the fraction:

[tex]\frac{\cos\ a\ cos\ b}{\cos\ a\cos b} - \frac{\sin a\ sin\ b}{\cos\ a\cos b} = 1 - \tan\ a\ tan\ b[/tex]

Cancel out common terms

[tex]1 - \frac{\sin a\ sin\ b}{\cos\ a\cos b} = 1 - \tan\ a\ tan\ b[/tex]

In trigonometry, we have:

[tex]\frac{\sin \theta}{\cos \theta} = \tan \theta[/tex]

So, the equation becomes:

[tex]1 - \tan\ a\ tan\ b = 1 - \tan\ a\ tan\ b[/tex]

Hence, the identity has been verified

Read more about trigonometry identities at:

https://brainly.com/question/21055284

Answer:

The interval 52% to 58% is based on a procedure that includes the true population value 95% of the time.

Step-by-step explanation:

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

95% confidence

We are 95% sure that the interval contains the true mean/proportion, and thus, the correct option is:

The interval 52% to 58% is based on a procedure that includes the true population value 95% of the time.

Answer:

is 7 pieces are remian

Step-by-step explanation:

9: total

2: eaten

so, 9-2 = 7 pieces?

Answer:

A. yes

Step-by-step explanation:

Answer:

team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.

Step-by-step explanation:

9514 1404 393

Answer:

  asymptotes: x = ±6

  zero: x = 0

Step-by-step explanation:

The vertical asymptotes of the function will be at the values of x where the denominator is zero. The denominator is x^2 -36, so has zeros for values of x that satisfy ...

  x^2 -36 = 0

  x^2 = 36

  x = ±√36 = ±6

The vertical asymptotes of the function are x = -6 and x = +6.

__

The zero of the function is at the value of x that makes the numerator zero. This will be the value of x that satisfies ...

  6x = 0

  x = 0 . . . . . divide by 6

The zero of the function is x=0.

__

As a check on this work, we have had a graphing calculator graph the function and identify the zero.

9514 1404 393

Answer:

Step-by-step explanation:

(a) As you might imagine, the disposition of apples in inventory will be one of "sold" or "discarded". (They could also be "stolen", but we'll call that "discarded", since they're not sold.) Then the inventory turnover T is the sum of numbers sold and discarded:

  T = s + d

__

(b) The value of d for Friday will be ...

  d = T -s = 34848 -29837 = 5,011 . . . value of d for Friday

Answer:

D

Step-by-step explanation:

x- coordinate = cos45° = [tex]\frac{1}{\sqrt{2} }[/tex] = [tex]\frac{\sqrt{2} }{2}[/tex]

y- coordinate = sin45° = [tex]\frac{1}{\sqrt{2} }[/tex] = [tex]\frac{\sqrt{2} }{2}[/tex]

That is ( [tex]\frac{\sqrt{2} }{2}[/tex] , [tex]\frac{\sqrt{2} }{2}[/tex] ) → D

Answer:

Step-by-step explanation:

Not a clear list of options and/or reference frame

Probably     0.5      if angle t is measured from the positive x axis.

Answer:

x-4y=2 can be written as y=(x-2)/4

(2,0) when x=2, y=0 and (6,1) when x=6, y=1

Answer:

[tex]A). \ \ \frac{(72\pi + 9\pi )}{4} \ in^2[/tex]

Step-by-step explanation:

Given;

radius of the circle, r = 9 inches

the part of the circle cut out = one-forth of the complete circle

the angle of the sector cut out θ= ¹/₄ x 360 = 90⁰

Area of the complete circle = πr² = π x 9² = 81π in²

Area of the sector cut out = [tex]= \frac{\theta }{360} \pi r^2 = \frac{90}{360} \pi (9^2) = \frac{1}{4} \times 81\pi = \frac{81 \pi}{4} = \frac{(72\pi + 9\pi)}{4} \ in^2[/tex]

Therefore, the only correct option is A. [tex]\frac{(72\pi + 9\pi )}{4} \ in^2[/tex]

We're told that "the claim that the population of student course evaluations has a mean equal to 3.50". So this means μ=3.50 makes up the null H0

The alternative would be H1: μ ≠ 3.50 since it's the opposite of the claim made in the null.

We go with answer choice D to form the null and alternative hypotheses.

The sign ≠ in the alternative hypothesis tell us that we have a two tail test.

---------------------------------------

Let's compute the test statistic

z = (xbar - mu)/(s/sqrt(n))

z = (3.44 - 3.50)/(0.67/sqrt(89))

z = -0.84483413122896

z = -0.84

The test statistic is roughly -0.84

---------------------------------------

Despite not knowing what sigma is (aka the population standard deviation), we can see that n > 30 is the case. So we can use the Z distribution. This is the standard normal distribution. When n > 30, the T distribution is fairly approximately the same as the Z distribution.

Use a calculator or a Z table to determine that

P(Z < -0.84) = 0.2005

which is approximate

Because we're doing a two-tail test, this means we double that result to get 2*0.2005 = 0.401

The p-value is roughly 0.401

-----------------------------------------

Since the p-value is larger than alpha = 0.05, we don't have enough evidence to reject the null. So you can say that we fail to reject the null, or we accept the null.

The conclusion based on that means that μ=3.50  must be true (unless other evidence comes along to disprove this). In other words, the mean evaluation score from students appears to be 3.50

Answer:

Answer 2/B

Step-by-step explanation:

The one with Parentheses

-(log6 x-4)

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\mathsf{= (3^{-1}+4^{-1})^{-2}}\\\\\\\mathsf{3^{-1} = \bf {\dfrac{1}{3}}}\\\\\\\mathsf{4^{-1} = \bf \dfrac{1}{4}}\\\\\\\mathsf{= (\dfrac{1}{3}+\dfrac{1}{4})^{-2}}\\\\\\\mathsf{\dfrac{1}{3} + \dfrac{1}{4} = \bf \dfrac{7}{12}}\\\\\\\mathsf{= (\dfrac{7}{12})^{-2}}\\\\\large\text{Simplify above and you have your overall answer...}\\\\\\\boxed{\boxed{\large\textsf{Answer: }\mathsf{\bf \dfrac{144}{49}}}}\huge\checkmark[/tex]

[tex]\huge\textsf{Good luck on your assignment \& enjoy your day!}\\\\\\\\\frak{Amphitrite1040:)}[/tex]

Answer:

The right response is Option c ($185,870,742).

Step-by-step explanation:

Given:

n = 30

r = 5.4%

or,

 = 0.054

Periodic payment will be:

[tex]R = \frac{360000000}{30}[/tex]

   [tex]=12000000[/tex] ($)

Now,

The present value will be:

= [tex]R+R(\frac{1-(1+r)^{-n+1}}{r} )[/tex]

By substituting the values, we get

= [tex]12000000+12000000(\frac{1-(1+0.054)^{-30 + 1}}{0.054} )[/tex]

= [tex]12000000+12000000\times 14.4892[/tex]

= [tex]185,870,742[/tex] ($)

Answer:

29/44

Step-by-step explanation:

[tex]\frac{4}{11} +\frac{5}{22} +\frac{3}{44} =\\[/tex]

-find the common denominator

[tex]\frac{4*4}{4*11} + \frac{2*5}{2*22} +\frac{3}{44} =[/tex]

[tex]\frac{16}{44} +\frac{10}{44} +\frac{3}{44} =[/tex]

-add the fractions and solve

[tex]\frac{16+10+3}{44} =[/tex]

[tex]\frac{29}{44}[/tex]

Answer:

CI 90% = ( 0.227 ; 0.273)

Step-by-step explanation:

Information from the survey:

sample size   n = 938

number of people with yes answer  x = 235

proportion of people  p = 235/938

p = 0.25    then  q = 1 - 0.25    q = 0.75

Confidence Interval 90 % .

CI 90% = ( p ± SE )

CI 90% = ( p ± z(c)*√(p*q)/n)

CI 90 %   then significance level is  α = 10 %   α/2 = 5%

α/2 = 0.05  we find in z-table z (c) = 1.64

√(p*q)/n  = √0.25*0.75/938

√(p*q)/n  = √0.000199

√(p*q)/n  = 0.014

CI 90% = ( p ± z(c)*√(p*q)/n)

CI 90% = ( 0.25 ± 1.64*0.014)

CI 90% = ( 0.25 ± 0.023 )

CI 90% = ( 0.227 ; 0.273)

Answer:

±√6

Step-by-step explanation:

[tex]15n=10(n+\frac{3}{n} )[/tex] is your expression first muiltiply out the 10 to get 15n= 10n+10 3/n next subtract  10 n from both sides to get 5n=10+3/n multiply both sides by n to get 5n^2=13 combine both sides and use the quadratic equation to solve to get your solution of ±√6

Solution :

a). The point estimate of proportion of college graduates among women who work at home,

[tex]$\hat p =\frac{166}{506}$[/tex]

  = 0.3281

99.5% confidence interval

[tex]$=\left( \hat p \pm Z_{0.005/2} \sqrt{\frac{\hat p (1- \hat p)}{n}} \right)$[/tex]

[tex]$=\left( 0.3281 \pm 2.81 \sqrt{\frac{0.3281 \times (1- 0.3281)}{506}} \right)$[/tex]

[tex]$=(0.3281 \pm 0.0586)$[/tex]

[tex]$=(0.2695, 0.3867)$[/tex]

Answer:

See below

Step-by-step explanation:

a)

[tex]2\sin(x) +\sqrt{3} =0 \implies 2\sin(x)=-\sqrt{3} \implies \boxed{\sin(x)=-\dfrac{\sqrt{3}}{2} }[/tex]

[tex]\therefore x=\dfrac{4\pi }{3}[/tex]

But note, as sine does represent the [tex]y[/tex] value, [tex]\dfrac{5\pi }{3}[/tex] is also solution

Therefore,

[tex]x=\dfrac{4\pi }{3} \text{ and } x=\dfrac{5\pi }{3}[/tex]

This is the solution for [tex]x\in[0, 2\pi ][/tex], recall the unit circle.

Note: [tex]\sin(x)=-\dfrac{\sqrt{3}}{2} \implies \sin(x)=\sin \left(\pi +\dfrac{\pi }{3} \right)[/tex]

b)

[tex]\sqrt{3} \tan(x) + 1 =0 \implies \tan(x) = -\dfrac{1}{\sqrt{3} } \implies \boxed{ \tan(x) = -\dfrac{\sqrt{3} }{3} }[/tex]

Once

[tex]\tan(x) = -\dfrac{\sqrt{3} }{3} \implies \sin(x) = -\dfrac{1}{2} \text{ and } \cos(x) = \dfrac{\sqrt{3} }{2}[/tex]

As [tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]

[tex]\therefore x=-\dfrac{\pi }{6}[/tex]

c)

[tex]4\sin^2(x) - 1 = 0 \implies \sin^2(x) = \dfrac{1}{4} \implies \boxed{\sin(x) = \pm \dfrac{\sqrt{1} }{\sqrt{4} } = \pm \dfrac{1}{2}}[/tex]

Therefore,

[tex]\sin(x)=\dfrac{1}{2} \implies x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6}[/tex]

[tex]\sin(x)=-\dfrac{1}{2} \implies x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]

The solutions are

[tex]x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6} \text{ and }x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]

Answer:          

35/5 (if you mean 10.6)    

1000000 (if you mean 10 to the sixth power)      

0.000001 (if you mean 10/6)

Answer:

There are 126 inches in 10'6

Step-by-step explanation:

take our feet and multiply the value by 12

Answer:

f = 10

g = 2 sqrt(3)

h = 20

Step-by-step explanation:

The short leg is opposite the smaller angle so it is f

The longer leg is opposite the larger angle so it is g

The hypotenuse is opposite the right angle so it is 20

We know f = x

g = x sqrt(3)

h = 2x = 20

2x = 20  so x = 10

f = 10

g = 2 sqrt(3)

h = 20

Answer:

The sample size needed is 3115.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

Point estimate:

[tex]\pi = 0.76[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

Calculate the sample size needed to be 95% confident that the error in estimating the true value of p is less than 0.015?

This is n for which M = 0.015. So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.015 = 1.96\sqrt{\frac{0.76*0.24}{n}}[/tex]

[tex]0.015\sqrt{n} = 1.96\sqrt{0.76*0.24}[/tex]

[tex]\sqrt{n} = \frac{1.96\sqrt{0.76*0.24}}{0.015}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.76*0.24}}{0.015})^2[/tex]

[tex]n = 3114.26[/tex]

Rounding up:

The sample size needed is 3115.