12
х
8
6
Find the value of x.
A) 9
B) 16
C) 14
D) 10

Answer:

(in the image)

Step-by-step explanation:

I'm not sure I understood your question completely but I hope this helps.

Answer:

(73.845 ; 76.155) ;

(41.633 ; 48.367) ;

1.273 ;

C. With 99% confidence, the mean width of all widgets is between 11.8 and 20.4. ;

1.717

Step-by-step explanation:

1.)

Given :

Mean, xbar = 75

Sample size, n = 24

Sample standard deviation, s = 3.3

α = 90%

Confidence interval = mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 90% ; df = 24 - 1 = 23

Tcritical = 1.714

Margin of Error = 1.714 * 3.3/√24 = 1.155

Confidence interval = 75 ± 1.155

Confidence interval = (73.845 ; 76.155)

2.)

Given :

Mean, xbar = 45

Sample size, n = 37

Sample standard deviation, s = 10.1

α = 95%

Confidence interval = mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 95% ; df = 37 - 1 = 36

Tcritical = 2.028

Margin of Error = 2.028 * 10.1/√37 = 3.367

Confidence interval = 45 ± 3.367

Confidence interval = (41.633 ; 48.367)

3.)

Given :

Mean, xbar = 37

Sample size, n = 30

Sample standard deviation, s = 4.1

α = 90%

Margin of Error = Tcritical * s/√n

Tcritical at 90% ; df = 30 - 1 = 29

Tcritical = 1.700

Margin of Error = 1.700 * 4.1/√30 = 1.273

5.)

Sample size, n = 23

Confidence level, = 90%

df = n - 1 ; 23 - 1 = 22

Tcritical(0.05, 22) = 1.717

Answer:

[tex] \frac{67}{30} \: \text{or} \:2 \frac{7}{30} [/tex]

Step-by-step explanation:

5/6 ÷ 1/3 - 2/3 (2/5)

Hope it helps you! ＼(^ᴥ^)／

9514 1404 393

Answer:

  66.5 cm²

Step-by-step explanation:

A horizontal line at the "knee" on the right will divide the figure into a 4 cm by 2 cm rectangle, and a trapezoid with bases 4 cm and 9 cm, and height 11-2 = 9 cm. Then the total area of the figure is ...

  A = LW + 1/2(b1 +b2)h

  A = (4 cm)(2 cm) + (1/2)(4 cm +9 cm)(9 cm) = 8 cm² +58.5 cm²

  A = 66.5 cm² . . . . area of the figure

Answer:

Interest rate is about 12.246%

The initial deposit doesn't matter because when you divide both sides by the initial deposit you're always left with (1+i)ⁿ=2

Step-by-step explanation:

[tex]4000(1+i)^6=8000\\(1+i)^6=2\\1+i=\sqrt[6]{2} \\1+i=1.122462048\\i=.12246[/tex]

Answer:

Price of X is $24.81

Price of Y is $3.66

Price of Z is $11.36

Step-by-step explanation:

for person A, we know that earns $40, then we can write the equation:

-4*z + 3*x + 3*y = $40

For person B, we know that earns $50, then:

1*z + 2*x - 3*y = $50

For person C, we know that earns $130, then:

6*z - 1*x + 4*y = $130

Then we have a system of equations:

-4*z + 3*x + 3*y = $40

1*z + 2*x - 3*y = $50

6*z - 1*x + 4*y = $130

To solve the system, we need to isolate one of the variables in one of the equations.

Let's isolate z in the second equation:

z = $50 - 2*x + 3*y

now we can replace this in the other two equations:

-4*z + 3*x + 3*y = $40

6*z - 1*x + 4*y = $130

So we get:

-4*($50 - 2*x + 3*y) + 3*x + 3*y = $40

6*($50 - 2*x + 3*y) - 1*x + 4*y = $130

Now we need to simplify both of these, so we get:

-$200 + 11x - 9y = $40

$350 - 13*x + 28*y = $130

Now again, we need to isolate one of the variables in one of the equations.

Let's isolate x in the first one:

-$200 + 11x - 9y = $40

11x - 9y = $40 + $200 = $240

11x = $240 + 9y

x = ($240 + 9y)/11

Now we can replace this in the other equation:

$350 - 13*x + 28*y = $130

$350 - 13*($240 + 9y)/11 + 28*y = $130

Now we can solve this for y.

- 13*($240 + 9y)/11 + 28*y  = $130 - $350 = -$220

-13*$240  - (13/11)*9y + 28y = - $220

y*(28 - (9*13/1) ) = -$220 + (13/11)*$240

y = ( (13/11)*$240 - $220)/(28 - (9*13/1) ) = $3.66

We know that:

x = ($240 + 9y)/11

Replacing the value of y, we get:

x = ($240 + 9*$3.66)/11  = $24.81

And the equation of z is:

z = $50 - 2*x + 3*y = $50 - 2* $24.81 + 3*$3.66 = $11.36

Then:

Price of X is $24.81

Price of Y is $3.66

Price of Z is $11.36

Answer:

x = 16

Step-by-step explanation:

The product of the lengths theorem is a property that can be sued to describe the relationships of the sides between the tangents and secants in a circle. One of these products states the following;

The distance between the point of tangency and its intersection point with the exterior secant squared is equal to the product of the exterior secant times the interior secant.

This essentially means the following equation can be formed;

[tex](AB)^2=(DC)(CB)[/tex]

Substitute,

[tex]12^2=x*9[/tex]

Simplify,

[tex]144=9x[/tex]

Inverse operations,

[tex]\frac{144}{9}=x\\\\16=x[/tex]

Answer:

[tex]\boxed{\sf x=7}[/tex]

Step-by-step explanation:

By Targent-secant theorem...

[tex]\sf 9(x + 9) = {12}^{2} [/tex]

Use the distributive property to multiply 9 by x+9.

[tex]\sf 9x+81= {12}^{2} [/tex]

Now, let calculate 12 to the power of 2 and get 144.

[tex]\sf 9x+81=144[/tex]

Subtract 81 from both sides.

[tex]\sf 9x=63[/tex]

Divide both sides by 9.

[tex] \sf \cfrac{ 9x}{9} = \cfrac{63}{9} [/tex]

[tex]\sf x=7[/tex]

Answer:

480 gallons.

Step-by-step explanation:

Given that in a mixture of 240 gallons, the ratio of ethanol and gasoline is 3: 1, if the ratio is to be 1: 3, to find the quantity of gasoline that is to be added the following calculation must be performed:

240/4 x 3 = Ethanol

240/4 = Gasoline

180 = Ethanol

60 = Gasoline

0.25 = 180

1 = X

180 / 0.25 = X

720 = X

720 - 180 - 60 = X

480 = X

Therefore, 480 gallons of gasoline must be added if the ratio is to be 1: 3.

Answer:

Step-by-step explanation:

Answer:

5 points per post and 2 points per replies

Answer:

hiiiiiii....!!! how r u

Answer:

Olha a foto.

Step-by-step explanation:

Answer:

v = ir

2 times 10 = 20v

Step-by-step explanation:

i think it is the one

Answer: x>1

Step-by-step explanation:

To solve for x, we want to isolate x.

3x+7>10       [subtract both sides by 7]

3x>3             [divide both sides by 3]

x>1

Therefore, we know that x>1.

Answer:

Step-by-step explanation:

3x + 7 > 10

3x > 10 - 7

3x > 3

x > 1

x ∈ ( 1, ∞ )

Answer:

7

Step-by-step explanation:

For simple interest,

I = prt

where I = interest,

p = principal (amount deposited)

r = annual rate of interest

t = time in years

We have r = 2% = 0.02

p = $3,000

I = $420

We need to find t

I = prt

420 = 3000 * 0.02 * t

420 = 60t

t = 420/60

t = 7

Answer: 7 years

The cordent plan of the answer is 2

Answer:

The scale factor will just be 2

Step-by-step explanation:

The length of PQ is twice as larger than the length of AB.

so from 12 to 6 or 6 to 12, we multiply 6 by 2 which equals to 12

Answer:

Bob had 25 cars

Step-by-step explanation:

10+15=25

Answer:

[tex]h=\sqrt{16(4)}[/tex]

[tex]h=8[/tex]

[tex]ANSWER:E) 8[/tex]

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

~HOPE IT HELPS

~HAVE A GREAT DAY!!

Answer:

B. Less than 1

Step-by-step explanation:

You could plug in values of n greater than 1 and see what happens....

Example n=2 gives |(.5+.2i)^2|

Simplifying inside gives |(.5)^2+2(.5)(.2i)+(.2i)^2|

=|.25+.2i+.04i^2|=|.25+.2i-.04|=|.21+.2i|.

Applying the absolute value part gives sqrt(.21^2+.2^2)=sqrt(.0441+.04)=sqrt(.0841)=.29

This value is less than 1.

We should also be able to do the absolute value first then the power.

|.5+.2i|=sqrt(.25+.04)=sqrt(.29)

So |.5+.2i|^2=.29 which is what we got long way around.

Anyways (sqrt(.29))^n where n is greater than 1 will result in a number greater than 0 but less than 1.

9514 1404 393

Answer:

  7.5

Step-by-step explanation:

Corresponding sides are proportional, so ...

  UV/VW = LM/MN

  x/6 = 15/12

  x = 6(15/12) = 15/2

  x = 7.5

Answer:

8d68d68fu9d6rf0d

c9yd7xpjd

puf68d6rif7

Answer:

50 bags ;

£750

Step-by-step explanation:

The dimension of the rectangular lawn is 500ft by 300 ft

The area of the lawn an e obtained thus :

Area of rectangle = Length * width

Area of rectangle = 500 ft * 300 ft

Area of rectangle = 150000 feets

1 bag of fertilizer covers 3000 feets

The minimum bags of fertilizer required :

Area of rectangle / Area covered by 1 bag of fertilizer

Minimum bags of fertilizer required :

(150,000 / 3000) = 50 bags

50 bags of fertilizer

Cost per bag = 15

Total cost = 15 * 50 = £750

Answer: 12

Step-by-step explanation:

The first step to dividing fractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. Finally, simplify the fractions if needed.

Answer:

See explanation

Step-by-step explanation:

Required

The average

The data whose average is to be calculated are not given.

However, the formula to calculate the average is:

[tex]\bar x = \frac{\sum x}{n}[/tex]

Assume the data is:

[tex]1287\ 1215\ 1221\ 1229[/tex]

This means that the number of schools is 4

So:

[tex]\bar x = \frac{1287+ 1215+ 1221+ 1229}{4}[/tex]

[tex]\bar x = \frac{4952}{4}[/tex]

[tex]\bar x = 1238[/tex]

The average of the assumed data is 1238

Answer:

9

Step-by-step explanation:

Let the number be X

From the problem we have the following equation:

7x - 17 = 3x + 19

-> 4x = 36

-> x = 9

Answer:

9

Step-by-step explanation:

that is the procedure above

Hello,

there are 5 differents sums:

16,18,20,22,24.

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

Dim i As Integer, j As Integer, k As Integer, l As Integer, u As Integer, v As Integer, nb As Integer

Dim mat(4, 4) As Integer

nb = 0

For i = 1 To 4

   For j = 1 To 4

       If j <> i Then

           For k = 1 To 4

               If k <> j And k <> i Then

                   l = 10 - k - j - i

                   If l > 0 And l < 5 And l <> i And l <> j And l <> k Then

                       mat(1, 1) = i

                       mat(1, 2) = j

                       mat(1, 3) = k

                       mat(1, 4) = l

                       For u = 2 To 4

                           For v = 1 To 4 - u + 1

                               mat(u, v) = mat(u - 1, v) + mat(u - 1, v + 1)

                           Next v

                       Next u

                       'Call visu(mat())

                       nb = nb + 1

                       Print nb,

                       mat(4, 1)

                   End If

               End If

           Next k

       End If

   Next j

Next i

End

Sub visu (m() As Integer)

   Dim i As Integer, j As Integer

   For i = 1 To 4

       For j = 1 To 4 - i + 1

           Print m(i, j);

       Next j

       Print

   Next i

End Sub

Answer:

(a)

[tex]Q_1 = 14[/tex]

[tex]Median = 15[/tex]

[tex]Q_3 = 20[/tex]

(b) [tex]IQR = 6[/tex]

Step-by-step explanation:

Given

[tex]14\ 14\ 15\ 26\ 13\ 16\ 21\ 20\ 15\ 13[/tex]

[tex]n = 10[/tex]

Solving (a): Median and the quartiles

Start by sorting the data

[tex]Sorted: 13\ 13\ 14\ 14\ 15\ 15\ 16\ 20\ 21\ 26[/tex]

The median position is:

[tex]Median = \frac{n + 1}{2}[/tex]

[tex]Median = \frac{10 + 1}{2} = \frac{11}{2} = 5.5th[/tex]

This implies that the median is the average of the 5th and the 6th data;

So;

[tex]Median = \frac{15+15}{2} = \frac{30}{2} = 15[/tex]

Split the dataset into two halves to get the quartiles

[tex]Lower: 13\ 13\ 14\ 14\ 15\[/tex]

[tex]Upper: 15\ 16\ 20\ 21\ 26[/tex]

The quartiles are the middle items of each half.

So:

[tex]Lower: 13\ 13\ 14\ 14\ 15\[/tex]

[tex]Q_1 = 14[/tex] ---- 14 is the middle item

[tex]Upper: 15\ 16\ 20\ 21\ 26[/tex]

[tex]Q_3 = 20[/tex] ---- 20 is the middle item

Solving (b): The interquartile range (IQR)

This is calculated as:

[tex]IQR = Q_3 - Q_1[/tex]

[tex]IQR = 20 - 14[/tex]

[tex]IQR = 6[/tex]

Solving (c): Incomplete details

[tex] \large{ \tt{❁ \: USING \: INTERNAL \: SECTION \: FORMULA: }}[/tex]

[tex] \large{ \bf{✾ \: P(x \:, y \: ) = ( \frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}} \: ,\: \frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}) }}[/tex]

[tex] \large{ \bf{⟹ \: ( \frac{8m + 3n}{m + n} , \: \frac{9m -n}{m + n}) }}[/tex]

[tex] \large{ \bf{ ⟼\frac{8m + 3n}{m + n} - \frac{9m - n}{m + n} - 2 = 0 }}[/tex]

[tex] \large{ \bf{⟼ \: \frac{8m + 3n - 9m + n}{m + n} - 2 = 0 }}[/tex]

[tex] \large{ \bf{⟼ \: \frac{4n - m}{ m + n} - 2 = 0 }}[/tex]

[tex] \large{⟼ \: \bf{ \frac{4n - m}{m + n }} = 2} [/tex]

[tex] \large{ \bf{⟼ \: 4n - m = 2m + 2n}}[/tex]

[tex] \large{ \bf{⟼ \: 4n -2 n = 2m + m}}[/tex]

[tex] \large{ \bf{⟼2n = 3m}}[/tex]

[tex] \large{ \bf{⟼ \: 3m = 2n}}[/tex]

[tex] \large{ \bf{⟼ \: \frac{m}{n} = \frac{2}{3} }}[/tex]

[tex] \boxed{ \large{ \bf{⟼ \: m : \: n = 2: \: }3}}[/tex]

-Hope I helped! Let me know if you have any questions regarding my answer and also notify me , if you need any other help! :)

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