Now suppose that bigger cups are ordered and the machine’s mean amount dispensed is set at μ=12. Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Given:

[tex]\to x^2+y^2=49[/tex]

When

 [tex]x=0\\\\0^2+y^2=49\\\\y^2=49\\\\y= \pm 7[/tex]

So, order pass [tex](0,\pm 7)[/tex]

Similarly When  

[tex]y=0\\\\x^2+0^2=49\\\\x^2=49\\\\x= \pm 7[/tex]

So, order pass [tex](\pm 7,0)[/tex]  

[tex]x \ \ \ \ \ \ \ 7 \ \ \ \ \ \ \ -7\ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ 0 \\\\y \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ 7 \ \ \ \ \ \ \ -7 \\\\[/tex]

Answer:

[tex]\sigma_x = 2000[/tex]

Step-by-step explanation:

Given

[tex]\sigma = 4000[/tex]

[tex]n = 4[/tex]

Required

The sample standard deviation

The sample standard deviation cannot be calculated directly. However, we can use the value of standard error because standard error approximates to sample standard deviation.

So, we have:

[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]

[tex]\sigma_x = \frac{4000}{\sqrt 4}[/tex]

[tex]\sigma_x = \frac{4000}{2}[/tex]

[tex]\sigma_x = 2000[/tex]

Hence, the sample standard deviation is 2000

I think

x= -4, -1 and y=13, 8

(-4, 13) and (-1, 8)

Answer:

35 ft^2

Step-by-step explanation:

A = wl

A = 7 * 5

A = 35

Answer:

.216165788

.582225057

.417774943

Step-by-step explanation:

We need to use a binomial distribution here

A.

10C5*.58⁵*(1-.58)⁵= .216165788

B.

I honestly think the fastest way to solve this is adding the probabiblity of exactly 6,7,8,9,10

which means we write

10C6*.58⁶*(1-.58)⁴+10C7*.58⁷*(1-.58)³+10C8*.58⁸*(1-.58)²+10C9*.58⁹*(1-.58)+10C10*.58¹⁰= .582225057

C.

To solve this just take the compliment of answer B

1-.582225057= .417774943

Step-by-step explanation:

what is your question please

Answer:

3.5h-3.1d-14

Step-by-step explanation:

6h+​(−8.1d​)−14+5d−2.5h

Combine like terms

6h -2.5h  -8.1d +5d  -14

3.5h-3.1d-14

Answer: GIVEN : f(x)= -3x+1

f(x)= -5

REQUIRE:  x=?

CALCULATION:

f(x)= -3x+1

As, given f(x)= -5.

Hence,

-5= -3x+1

OR

-3x+1= -5

-3x= -5-1

-3x=-6

x= -6/-3

x= 2

Answer:

√53

Step-by-step explanation:

Distance between two points =  

√(4−6)^2+(−2−5)^2

√(−2)^2+(−7)^2  

= √4+49

=√53

= 7.2801

Hope this helps uwu

9514 1404 393

Answer:

  option 2: √53

Step-by-step explanation:

The distance formula is useful for this:

  d = √((x2 -x1)² +(y2 -y1)²)

  d = √((4-6)² +(-2-5)²) = √((-2)² +(-7)²) = √(4+49)

  d = √53

The distance between the given points is √53.

Answer:

84 sq meters

Step-by-step explanation:

1. Approach

In order to solve this problem, one will have to divide the figure up into simple shapes. A picture is attached showing how the shape is divided up for this answer. Find the area of each region, then add up the results to find the total area.

2. Area of Region 1

As one can see, the length of (Region 1), as given is (6), the width is (3). To find the area multiply the length by the width.

Length * width

6 * 3

= 18

3. Area of Region 2

In (Region 2), the length is given, (12). However, one must find the width, this would be the size of the total side, minus the width of (Region 1). Multiply the length by the side to find the area.

Length * width

= 12 * (8 - 3)

= 12 * 5

= 60

4. Area of Region 3

In (Region 3), the length of the figure is (2), the width is (3). To find the area, multiply the length by the width.

Length * width

= 2 * 3

= 6

5. Total area

Now add up the area of each region to find the total rea,

(Region 1) + (Region 2) + ( Region 3)

= 18 + 60 + 6

= 84

Answer:

a) n and o

Step-by-step explanation:

by the concept of corresponding angles, angles that are equal when a line intersects 2 parallel lines .

l is the line that intersects n and o, and since x = 35, theyre corresponding angles resulting in n and o being parallel

Answer:

a) n and o

Step-by-step explanation:

Corresponding angles

9514 1404 393

Answer:

  C = 9x +2280

Step-by-step explanation:

The 2-point form of the equation for a line can be a place to start.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  C = (4800 -3000)/(280 -80)(x -80) +3000 . . . . . use given values

  C = 1800/200(x -80) +3000 . . . . simplify some

  C = 9x -720 +3000 . . . . eliminate parentheses

  C = 9x +2280

9514 1404 393

Answer:

  (9/34)√34 ≈ 1.543

Step-by-step explanation:

The second equation can be rewritten as ...

  6x -10y -12 = 0

  3x -5y -6 = 0

  3x -5y = 6

__

The formula for the distance from point (x, y) to line ax+by+c=0 is ...

  d = |ax+by+c|/√(a²+b²)

Then the distance from a point to the first line is ...

  d = |3x -5y +3|/√(3² +(-5)²)

We know from the rearrangement of the second equation that points on its line satisfy (3x-5y) = 6. Substituting this value for (3x -5y) in the distance formula gives ...

  d = |6 +3|/√34

Simplifying and rationalizing the denominator gives a distance of ...

  d = (9/34)√34 ≈ 1.543

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

[tex]\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right][/tex]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Answer:

B

Step-by-step explanation:

Find the area of the circle with r = 36/2     Area1 =  π r²

divide Area1 by two since the upper part of the figure is a semi-circle

then finally add and area of the rectangle   Area2 = (18)(36)

Total area = Area 1 + Area 2

Answer:

1156.7 cm^2

Step-by-step explanation:

for finding area of semi circle

diameter=36 cm

radius=diameter/2

=36/2

=18 cm

area of semi circle=πr^2/2

=3.14*(18)^2/2

=3.14*324/2

=1017.36/2

=508.68 cm^2

area of rectangle=length*breadth

=18 cm *36 cm

=648 cm^2

area of the figure=area of semi circle + area of rectangle

=508.68 + 648

=1156.68

=1156.7 cm^2

Hello~

I used Cymath for this!

https://www.cymath.com/answer?q=(0.4%20*10%5E-6)%20(0.7%20*%2010%5E-2)

In short the answer is B.

The link has the step by step answers!

I highly recommend this sight by the way, its always correct!

Ary~

Answer:

0.64(x) = 30

Step-by-step explanation:

Hope that's correct.

Answer:

Step-by-step explanation:

1)

[tex]cot x sec^4 x = cotx + 2tanx + tan^3 x[/tex]

[tex]RHS =[/tex]

           [tex]=\frac{cosx}{sinx} + 2 \frac{sinx }{cosx} + \frac{sin^3x }{cos^3x}\\\\[/tex]

          [tex]= \frac{cosx(cos^3x) + 2sinx(sinx \ cos^2x ) + sin^3x(sinx)}{sinx \ cos^3x}\\\\[/tex]        [tex][\ taking LCM \ ][/tex]

         [tex]=\frac{cos^4x + 2sin^2xcos^2x +sin^4x }{sinx cos^3x}[/tex]

         [tex]= \frac{(sin^2 x + cos^2x)^2 }{sin x \ cos^3 x}[/tex]                        [tex][ \ a^4 + 2a^2b^2 + b^4 = (a^2 + b^2 ) ^2 \ ][/tex]

        [tex]= \frac{1}{sinx \ cos^3 x}\\\\= \frac{1 \times cosx}{sinx \times cos^3x \times cosx}[/tex]                    [tex][ \ multiplying\ and \ dividing \ by \ cosx \ ][/tex]

        [tex]= \frac{cosx}{sinx} \times \frac{1}{cos^4x}\\\\=cot x \ sec^4 x[/tex]

        [tex]= LHS[/tex]

2)

[tex]sin x \ ( tanx \ cosx - cotx \ cosx) = 1 - 2cos^2x[/tex]

[tex]LHS =[/tex]

        [tex]=sinx( [ \frac{sinx}{cosx} \times cosx)] - [ \frac{cosx}{sinx}\times cosx] )\\\\= sinx (sinx - \frac{cos^2x}{sinx})\\\\=sin^2x - cos^2 x\\\\=(1 - cos^2x ) -cos^2 x[/tex] [tex][ \ sin^2x = 1 -cos^2 x \ ][/tex]

        [tex]= 1 -cos^2x - cos^2 x \\\\= 1 - 2cos^2x \\\\=RHS[/tex]

3)

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

[tex]LHS =[/tex]

       [tex]= 1 +sec^2x \sin^2x \\\\= 1 + (\frac{1}{cos^2x} \times sin^2x )\\\\= 1 + \frac{sin^2 x}{cos^2x}\\\\= 1 + tan^2x \\\\= sec^2 x\\\\=RHS[/tex]

4)

[tex]\frac{sinx}{1 -cosx} + \frac{sinx}{1+cosx} = 2 \ cosec x[/tex]

[tex]LHS =[/tex]

[tex]=\frac{sinx}{1 -cosx} + \frac{sinx}{1+cosx} \\\\= \frac{sinx(1 +cosx)}{(1-cosx)(1+cosx)} + \frac{sinx(1-cox)}{(1+cosx)(1-cosx)}\\\\= \frac{sinx +sinx\ cosx}{(1 - cos^2x)} + \frac{sinx - sinx \ cosx}{1 - cos^2x}\\\\=\frac{sinx + sinx \ cosx + sinx - sinx \ cosx}{1 - cos^2x}\\\\=\frac{2sinx}{sin^2x}\\\\=\frac{2}{sinx}\\\\=2 cosec\ x\\\\=RHS[/tex]

5)

[tex]- tan^2 + sec^2 x = 1\\\\[/tex]

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

Answer:

[tex]y = -\frac{1}{3}m + 41[/tex]

Step-by-step explanation:

Linear equation:

A linear function has the following format:

[tex]y = mn + b[/tex]

In which m is the slope and b is the y-intercept.

When she plants 30 stalks, each plant yields 31 oz of beans. When she plants 36 stalks, each plant produces 29 oz of beans.

This means that these two points belong to the line: (30,31), (36,29).

Finding the slope:

When we have two points, the slope is given by the change in the output divided by the change in the input.

Change in the output: 29 - 31 = -2

Change in the input: 36 - 30 = 6

Slope:

[tex]m = \frac{-2}{6} = -\frac{1}{3}[/tex]

Thus

[tex]y = -\frac{1}{3}m + b[/tex]

Finding b:

We take one of the points and replace on the equation.

(30,31) means that [tex]m = 30, y = 31[/tex]. Thus

[tex]y = -\frac{1}{3}m + b[/tex]

[tex]31 = -\frac{1}{3}(30) + b[/tex]

[tex]31 = -10 + b[/tex]

[tex]b = 41[/tex]

Thus

[tex]y = -\frac{1}{3}m + 41[/tex]

220 cubic units.

Answer:

Solution given:

for small cylinder

r=1

and for large cylinder

R=5+1=6

height for both [h]=2

Now

Volume of solid=πR²h-πr²h=πh(R²-r²)

=3.14*2(6²-1²)=219.8 =220 units ³.

Small cylinder is r=1

Large cylinder is R= 5+1 =6

Height (h) =2

Volume of solid,

→ πR²h-πr²h

→ πh(R²-r²)

→ 3.14 × 2(6²-1²)

→ 219.8

→ 220 cubic units

Answer:

Your measurements; Area = 216.108 cm²

Another student's measurements; Area = 216.9404 cm²

- Difference in area could be as a result of human error or perhaps that they made use of different measuring tools.

Step-by-step explanation:

For Your measurements;

Length of rectangle = 20.70 cm

Width of rectangle = 10.44 cm

Area of rectangle is given by; A = length × width = 20.7 × 10.44 = 216.108 cm²

For Another student's measurements;

Length of rectangle = 20.74 cm

Width of rectangle = 10.46 cm

Area = 20.74 × 10.46

Area = 216.9404 cm²

The areas they both obtained are not of equal values and this could be as a result of human error or perhaps that they used different measuring tools.

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is [tex]\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}[/tex].

Now we will solve this expression with the help of law of exponents.

[tex]\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}[/tex]

           [tex]=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}[/tex]

           [tex]=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}[/tex]

           [tex]=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}[/tex]

           [tex]=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}[/tex]

           [tex]=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}[/tex]

           [tex]=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }[/tex] [Option 2]

[tex]2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2[/tex] [Option 1]

[tex]2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2[/tex]

                [tex]=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }[/tex]

                [tex]=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }[/tex] [Option 3]

[tex]2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }[/tex]

               [tex]=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}[/tex] [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

Answer:

The answer is 7,953.

Step-by-step explanation :

When you divide like this you are basically being asked how many times something can go into something else. In this case, we are being asked to calculate how many times 1000 can go into 7,953,000. You can use basic math skills or do long division.

I'm going to go with the easier way, basic math skills.

There are three 0's in the back of the number 7,953,000, and notice 1000 has three 0's as well in the back. When you divide those three 0's in the back of both numbers basically cancel out. Well, what does that leave us with?

7953 is what were left with. The answer is 7,953.

Answer:

Area of given figure = 60.8 mm² (Approx.)

Step-by-step explanation:

Given:

Length of figure = 9.3 mm

Width of figure = 7.8 mm

Height of triangular cut = 3 mm

Find:

Area of given figure

Computation:

Area of given figure = [(Length)(Width)] - Area of triangle

Area of given figure = [(9.3)(7.8)] - [(1/2)(b)(h)]

Area of given figure = [(9.3)(7.8)] - [(1/2)(7.8)(3)]

Area of given figure = [(9.3)(7.8)] - [(1/2)(7.8)(3)]

Area of given figure = [72.54] - [(1/2)(7.8)(3)]

Area of given figure = [72.54] - [11.7]

Area of given figure = 60.8 mm² (Approx.)