find the greatest number than divides 45 60 75 without leaving remainder​

Answer:

1/8 is less than

Step-by-step explanation:

i dont see any fractions below gona have to edit your answer

Answer:

$80

Step-by-step explanation:

To find the largest price, assume that all 20 copies of the workbook will have 400 pages.

Since the company charges 1 cent per page, this means each workbook will cost 400 cents. This is equivalent to 4 dollars.

Find the total cost by multiplying this by 20:

20(4)

= 80

So, the largest price to print 20 copies is $80

Answer:

25.40

Step-by-step explanation:

tickets  ( 2 at 10.95 each) = 2* 10.95 = 21.90

popcorn ( 1 at 7.50)         = 7.50

Total cost before discount

21.90+7.50=29.40

subtract the discount

29.40-4.00 =25.40

Answer:

Yep! That's correct!

Step-by-step explanation:

We know that Marilyn and her sister are each getting a ticket that cost $10.95. They are also getting a $7.50 popcorn to share. Let's add those values up.

(10.95 * 2) + 7.50 {Multiply 10.95 by 2 to get 21.90.}

21.90 + 7.50 {Add 7.50 to 21.90 to get 29.40}

$29.40 (without the credit) in toal

A credit on a movie reward card functions as a discount, so what we need to do next is subtract 4 from 29.40. That will get us $25.40 as the total cost.

After doing the math, I can deduce that your answer is correct!

Answer:

The angles are 45 and 135

Step-by-step explanation:

The two angles form a straight line, which is 180 degrees

c+ 3c = 180

4c = 180

Divide by 4

4c/4 =180/4

c = 45

3c = 3(45) = 135

The angles are 45 and 135

Answer:

45 and 135 ...

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

Answer:

279+x

Step-by-step explanation:

Emily + Yani + Joyce=3209 stickers

how many stickers does Emily have than Joyce:

(279+2x)-(x)

279+2x-x

=279+x

Answer:

X =224

Y= -10

Step-by-step explanation:

To solve this question it's better to convert the fractions to decimals this way it will be easy to solve.

0.25x+0.6y= -4

0.2x+0.25y=-0.9

0.2(0.25x+0.6y=-4)

0.25(0.2x+0.25y=-0.9)

0.05x+0.12y=-0.8

0.05x+0.06y=-0.225

0.0575y/0.0575=-0.575/0.0575

Y=-10

To find x you replace the value of y in any of the equations

0.25x+0.6y=-4

0.25x+0.6(-10)=-4

0.25x=-4+60

0.25x/0.25=56/0.25

X=224

I hope this helps and sorry if it's wrong

volume= a^2 * h

area= a^2+4ah

take the second equation, solve for h

4ah=1100-a^2

h=1100/4a -1/4 a now put that expression in volume equation for h.

YOu now have a volume expression as function of a.

take the derivative, set to zero, solve for a. Then put that value back into the volume equation, solve for Volume.

Cited from jiskha

Answer:

F statistic = 2.124

Pvalue = 0.0546

A. Fail to reject the null hypothesis. There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

Step-by-step explanation:

H0 : pain reduction is the same

H1 : pain reduction is varies more with sham.

Sham n= 20 x=0.41 s=1.37

Magnet n= 20 x =0.46 s= 0.94

α - level = 0.05

Using the Ftest statistic

Ftest = larger sample variance / smaller sample variance

Ftest = s1² / s2² = 1.37² / 0.94² = 1.8769 / 0.8836 = 2.124

The degree of freedom :

Numerator = n - 1 = 20 - 1 = 19

Denominator = n - 1 = 20 - 1 = 19

Pvalue(2.124, 19, 19) = 0.0546

Since ;

Pvalue > α ; WE fail to reject the Null ; Result is not significant

Answer:

I think you can go with:

The margin of error is equal to half the width of the entire confidence interval.

so  try .74 ±   =   [ .724 , .756] as the confidence interval

Step-by-step explanation:

this the answer of queastions

Step-by-step explanation:

67.18,68.82

Let mu be the true population mean of statistics exam scores. We have a large random samples of n=36 scores with a sample mean of 68.we know that the population standard deviation is sigma=3.A pivotal quantity is 3^sqrt(36)=(3/6)=68(1/2) which is approximately normally distributed. Therefore the 85%confidence interval is 68-(1/2)(1.6449), 68+(1/2)(1.6449) i.e (67.18,68.82)

Answer:

Step-by-step explanation:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

Answer:

32

Step-by-step explanation:

3(8(3)-5(2))-2((3)+(2))

3(24-10) -2(5)

3(14) -10

42-10

32

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

[tex]\huge\textsf{3(8a - 5b) - 2(a + b)}\\\\\huge\textsf{= 3(8(3) - 5(2)) - 2(3 + 2)}\\\\\huge\textsf{= 3(24 - 10) - 2(3 + 2)}\\\\\huge\textsf{= (3)(14) - 2(3 + 2)}\\\\\huge\textsf{= 42 - 2(3 + 2)}\\\\\huge\textsf{= 42 - 2(5)}\\\\\huge\textsf{= 42 - 10}\\\\\huge\textsf{= 32}}[/tex]

[tex]\huge\boxed{\textsf{Answer: 32}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\huge\boxed{\frak{Amphitrite1040:)}}[/tex]

9514 1404 393

Answer:

  (-2, 4]

Step-by-step explanation:

  -21 ≤ -6x +3 < 15 . . . . given

  -24 ≤ -6x < 12 . . . . . . subtract 3

  4 ≥ x > -2 . . . . . . . . . . divide by -6

In interval notation, the solution is (-2, 4].

__

Interval notation uses a square bracket to indicate the "or equal to" case--where the end point is included in the interval. A graph uses a solid dot for the same purpose. When the interval does not include the end point, a round bracket (parenthesis) or an open dot are used.

Answer:

H0 : correlation is equal to 0

H1 : correlation is not equal to 0 ;

Pvalue < α ;

There is sufficient evidence

r = 0.945 ;

Pvalue = 0.01524

Step-by-step explanation:

Given the data :

Lemon_Imports_(x) Crash_Fatality_Rate_(y)

230 15.8

264 15.6

359 15.5

482 15.3

531 14.9

Using technology :

The regression equation obtained is :

y = 16.3363-0.002455X

Where, slope = - 0.002455 ; Intercept = 16.3363

The Correlation Coefficient, r = 0.945

H0 : correlation is equal to 0

H1 : correlation is not equal to 0 ;

The test statistic, T:

T = r / √(1 - r²) / (n - 2)

n = 5 ;

T = 0.945 / √(1 - 0.945²) / (5 - 2)

T = 0.945 / 0.1888341

T = 5.00439

The Pvalue = 0.01524

Since Pvalue < α ; Reject the Null and conclude that there is sufficient evidence to support the claim.

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

Answer:slope 2/3

Y-int 6

Step-by-step explanation:

Answer:

-7/25

Step-by-step explanation:

1/5 ÷ (-5/7)

Copy dot flip

1/5 * -7/5

-7/25

Answer:

I think it is twenty seven

Answer:

x = 3

Step-by-step explanation:

7(x + 1) + 2 = 5x + 15

~Simplify left side

7x + 7 + 2 = 5x + 15

~Combine like terms

7x + 9 = 5x + 15

~Subtract 9 to both sides

7x = 5x + 6

~Subtract 5x to both sides

2x = 6

~Divide 2 to both sides

x = 3

Best of Luck!

Recall that

[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

Differentiating the power series series for y(x) gives the series for y'(x) :

[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

Now, replace everything in the DE with the corresponding power series:

[tex]y'-2xy = 6\sin(3x) \implies[/tex]

[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.

Split up both series on the left into even- and odd-indexed series:

[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]

[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]

Next, we want to condense the even and odd series:

• Even:

[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]

• Odd:

[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].

The even series vanishes, so that

[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]

for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find

[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]

[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]

and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].

This leaves us with the odd series,

[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]

[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]

We have

[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]

[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]

[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]

[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]

So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then

[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]

[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]

[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]

and so the first four terms of series solution to the DE would be

[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]

Answer

nghiệmTrảingu       từng bước:

Answer:

A

Step-by-step explanation:

f(-1)=7, f(0)=2, f(1)=-3

9514 1404 393

Answer:

  69.1 ft

Step-by-step explanation:

The diameter of the circle is 24 ft. The length of the arc is more than twice the diameter, so cannot be less than about 50 ft. The only reasonable choice is ...

  69.1 ft

__

The circumference of the circle is ...

  C = 2πr = 2(3.14)(12 ft) = 75.36 ft

The arc length of interest is 330° of the 360° circle, so is 330/360 = 11/12 times the circumference.

  s = (11/12)(75.36 ft) = 69.08 ft ≈ 69.1 ft

Answer:D

Step-by-step explanation:

Answer: -2(d) is the answer.

Step-by-step explanation:

x1 = 3

y1 = -5

x2 = -2

y2 = 5

slope (m) = rise/run = (y2 - y1)/(x2-x1)

                                =(5-(-5))/(-2-3)

                                 =   10/-5

                                 = -2

9514 1404 393

Answer:

  B. 11.000

Step-by-step explanation:

The function looks like a reflected and translated exponential function with a horizontal asymptote near y = 11.000. The rate of change is decreasing so fast that the next value is expected to be very near 10.994. The closest one among the answer choices is 11.000.

_____

First differences are 0.811 and 0.011. The latter is about 0.0136 times the former. At that rate of change, we expect the next first difference to be about 0.000149, which would make the next number in sequence be about 10.9941—very little change from 10.994.

Clearly, first differences are not constant, so the function is not linear. Ratios of the numbers are not constant, so this is not an exponential (geometric) sequence. A reflected exponential function of the type described is a good fit.

With only 3 points given, the rule is not at all obvious. The next term could legitimately be anything you like, and a rule could be made that would fit it.

Answer:

Consider the following identity:

Let a = 2, b = 1/2

Use the algebraic identity given below

[tex]\boxed{\sf a^3-b^3=(a+b)(a^2-ab+b^2)}[/tex]

[tex]\\ \sf\longmapsto (2+\dfrac{1}{2})(2^2-1+\dfrac{1}{4})[/tex]

[tex]\\ \sf\longmapsto (2+\dfrac{1}{2})(2^2-2\times \dfrac{1}{2}+\dfrac{1}{2}^2)[/tex]

[tex]\\ \sf\longmapsto 2^3-\dfrac{1}{2}^3[/tex]

[tex]\\ \sf\longmapsto 8-\dfrac{1}{8}[/tex]

Answers:

First bottle's unit cost = 27 cents per oz

Second bottle's unit cost = 27 cents per oz

Both have the same unit cost.

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

Work Shown:

unit cost = price/(number of ounces)

1st bottle unit cost = (1.35)/(5) = 0.27 dollars per oz = 27 cents per oz

2nd bottle unit cost = (2.16)/(8) = 0.27 dollars per oz = 27 cents per oz

Both lead to the same unit cost. Therefore, you can pick either option and it doesn't matter.

Answer:

Oh no I am sorry! If you want answers to be done the real way let me know

Answer:I'm so sorry for you but congrats you did get the answer right it's just the test I guess

Step-by-step explanation: