(07.04 LC)What is the solution to the equation ax = 2?
Ox= 1
Ox=1
- کی
Ox= 3
O x = 12
Plz helppp

Answer:

[tex]-3 < x[/tex]

Step-by-step explanation:

Given

[tex]-18 < 5x-3[/tex]

Required

The solution

[tex]-18 < 5x-3[/tex]

Add 3 to both sides

[tex]3-18 < 5x-3+3[/tex]

[tex]-15 < 5x[/tex]

Divide both sides by 5

[tex]-3 < x[/tex]

Answer:

6 hrs 33 minutes

Step-by-step explanation:

 2hr 57min

+3hrs42min​

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

5 hrs 99 minutes

But 1 hr = 60 minutes so subtract 60 minutes and add 1 hour

6 hrs 33 minutes

Answer:

6hr 39 min

Step-by-step explanation:

Add both

2hr 57 min

+ 3hr 42 min

5hr 99 min

we know that 1 hr = 60 min

then , 99 min = 1hr 39 min

so, 5hr + 1hr 39min

= 6hr 39 min

Answer:

The solution is D. x² + y² + 4x - 2y = -1

Step-by-step explanation:

The standard form of a circle with a center at (h,k) and a radius r is:

(x-h)² + (y-k)² = r²

Since the center is (-2,1) and the radius is

2, we know that:

Thus, the equation of the circle is:

(x-(-2))² + (y-1)² = 2²

This simplifies to be

(x+2)² + (y-1)² = 4

The equation of the circle is:

(x+2)² + (y-1)² = 4

x²+4x+4+y²-2y+1 = 4

x²+y²+4x-2y+5 = 4

x²+y²+4x-2y = 4-5

x²+y²+4x-2y = -1

Answer:

try all the square roots and wichever gets to 0.58(rounded) is your answer

Step-by-step explanation:

Answer:

x = 7.348

Step-by-step explanation:

I don't really know how to explain this over a computer, but it is correct. Answer credit of Omnicalculator.

Hope that this helps!

Answer:

x=41

Step-by-step explanation:

LM =JM

154=4x-10

154+10=4x

164=4x

164/4=4x/4

41=x

hope this is helpful

Answer:

x+4=72

x=72-4

x=68

Step-by-step explanation:

let x be one number and 4 be anothe number then sum of the two number be x and 4

Answer:

Scale factor = ½

Step-by-step explanation:

The original image is the preimage = ∆ACB

The new image is the image = ∆DFE

The scale of factor of dilation = image/preimage = DF/AC

DF = 6 cm (given)

AC = 12 cm (given)

Plug in the values into the equation to find the scale factor of dilation:

Scale factor = 6/12

Scale factor = ½

9514 1404 393

Answer:

  3.464

Step-by-step explanation:

Comparing the given equation to the standard form equation of a circle:

  (x -h)^2 +(y -k)^2 = r^2

we find the center to be (h, k) = (0, 1) and the radius to be √12 = 2√3.

The problem statement tells us to round the radius to the nearest thousandth, so it will be ...

  2√3 ≈ 3.4641016 ≈ 3.464 . . . radius of the circle

Answer:

A will have this new coordinate (2, 1).

S will have this new coordinate (-2, 7).

E will have this new coordinate (2, 7).

Step-by-step explanation:

A is the coordinate (-1, 2).

S is the coordinate (-7, -2).

E is the coordinate (-7, 2).

To rotate that figure 270 grades (is the same to rotate 90 grades) we need to convert those coordinates to this (y, -x).

So,

A will have this new coordinate (2, 1).

S will have this new coordinate (-2, 7).

E will have this new coordinate (2, 7).

Answer:

0.2 amps

Step-by-step explanation:

Given data

The formula V=IR is the formula for ohms law

Which state that  the voltage is directly proportional to the current and the resistance in an electric circuit

Now

R= 15 ohms

V= 3volts

V= IR

3= I*15

I= 3/15

I= 0.2 amps

Hence he current flowing is 0.2 amps

Answer:

(a):

Dannette                   Alphonso

[tex]\bar x_D = 4.33[/tex]                    [tex]\bar x_A = 5.17[/tex]

[tex]M_D = 2.5[/tex]                    [tex]M_A = 5[/tex]

[tex]\sigma_D = 3.350[/tex]                  [tex]\sigma_A = 1.951[/tex]

[tex]IQR_D = 7[/tex]                  [tex]IQR_A = 1.5[/tex]

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

[tex]A = \{3,4,4,4,4,5,5,5,5,6,6,11\}[/tex] ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

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

So, we have:

[tex]\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}[/tex]

[tex]\bar x_D = \frac{52}{12}[/tex]

[tex]\bar x_D = 4.33[/tex] --- Dannette

[tex]\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}[/tex]

[tex]\bar x_A = \frac{62}{12}[/tex]

[tex]\bar x_A = 5.17[/tex]  --- Alphonso

Median

The median is calculated as:

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

[tex]M = \frac{12 + 1}{2}th[/tex]

[tex]M = \frac{13}{2}th[/tex]

[tex]M = 6.5th[/tex]

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

[tex]M_D = \frac{2+3}{2}[/tex]

[tex]M_D = \frac{5}{2}[/tex]

[tex]M_D = 2.5[/tex] ---- Dannette

[tex]M_A = \frac{5+5}{2}[/tex]

[tex]M_A = \frac{10}{2}[/tex]

[tex]M_A = 5[/tex]  ---- Alphonso

Standard Deviation

This is calculated as:

[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]

So, we have:

[tex]\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}[/tex]

[tex]\sigma_D = \sqrt{\frac{134.6668}{12}}[/tex]

[tex]\sigma_D = 3.350[/tex] ---- Dannette

[tex]\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}[/tex]

[tex]\sigma_A = \sqrt{\frac{45.6668}{12}}[/tex]

[tex]\sigma_A = 1.951[/tex] --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

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

Where

[tex]Q_3 \to[/tex] Upper Quartile       and        [tex]Q_1 \to[/tex] Lower Quartile

[tex]Q_3[/tex] is calculated as:

[tex]Q_3 = \frac{3}{4}*({n + 1})th[/tex]

[tex]Q_3 = \frac{3}{4}*(12 + 1})th[/tex]

[tex]Q_3 = \frac{3}{4}*13th[/tex]

[tex]Q_3 = 9.75th[/tex]

This means that [tex]Q_3[/tex] is the mean of the 9th and 7th item. So, we have:

[tex]Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16[/tex]           [tex]Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11[/tex]

[tex]Q_3 = 8[/tex] ---- Dannette                 [tex]Q_3 = 5.5[/tex] --- Alphonso

[tex]Q_1[/tex] is calculated as:

[tex]Q_1 = \frac{1}{4}*({n + 1})th[/tex]

[tex]Q_1 = \frac{1}{4}*({12 + 1})th[/tex]

[tex]Q_1 = \frac{1}{4}*13th[/tex]

[tex]Q_1 = 3.25th[/tex]

This means that [tex]Q_1[/tex] is the mean of the 3rd and 4th item. So, we have:

[tex]Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2[/tex]                  [tex]Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8[/tex]

[tex]Q_1 = 1[/tex] --- Dannette                   [tex]Q_1 = 4[/tex] ---- Alphonso

So, the IQR is:

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

[tex]IQR_D = 8 - 1[/tex]                                     [tex]IQR_A = 5.5 - 4[/tex]

[tex]IQR_D = 7[/tex] --- Dannette                      [tex]IQR_A = 1.5[/tex] --- Alphonso

Solving (b): The measures to compare

Measure of  center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of  spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

[tex]Lower =Q_1 - 1.5 * IQR[/tex]

[tex]Upper =Q_3 + 1.5 * IQR[/tex]

For Dannette:

[tex]Lower = 1 - 1.5 * 7 = -9.5[/tex]

[tex]Upper = 8 + 1.5 * 7 = 18.5[/tex]

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

[tex]Valid = 0 \to 18.5[/tex]

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

[tex]Lower = 4 - 1.5 * 1.5 =1.75[/tex]

[tex]Upper = 5.5 + 1.5 * 1.5 =7.75[/tex]  

So, the valid data range are:

[tex]Valid = 1.75\to 7.75[/tex]

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]y=-\frac{1}{5}x+\frac{23}{5}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{\underline{Slope Is...}}}\\\\\rightarrow\frac{y_2-y_1}{x_2-x_1}\\\\\boxed{\text{Key:}}\\\\\rightarrow (x_1,y_1)\text{ and }(x_2,y_2) - \text{Two Points Given}[/tex]

⸻⸻⸻⸻

[tex]\boxed{\text{Calculating the Slope...}}\\\\\rightarrow m=\frac{4-5}{3-(-2)}\\\\\rightarrow m=\frac{-1}{5}\\\\\rightarrow \boxed{m=-\frac{1}{5}}\\\\\\\text{The slope is } -\frac{1}{5}.[/tex]

⸻⸻⸻⸻

⸻⸻⸻⸻

[tex]\boxed{\text{Calculating the intercept...}}\\\\y=-\frac{1}{5}x+b\\-------------\\\rightarrow 4 = -\frac{1}{5}(3) +b\\\\\rightarrow4= -\frac{3}{5}+b\\\\\rightarrow4+\frac{3}{5}= -\frac{3}{5}+\frac{3}{5} +b\\\\\rightarrow \boxed{\frac{23}{5}=b}[/tex]

⸻⸻⸻⸻

[tex]\text{The equation should be: }\boxed{ y=-\frac{1}{5}x+\frac{23}{5} }.[/tex]

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

I don’t know what type of figure you are talking about. But if there is supposed to be a picture attached, we see nothing

Step-by-step explanation:

You could repost this question though, I’ll be glad to help on that other question!

Answer:

y = 30.96

Step-by-step explanation:

take 23 degree as reference angle

using tan rule

tan 23 = opposite / adjacent

0.42 = 13/y

y = 13/0.42

y = 30.96

Answer:

The t-ratio is the estimate divided by the standard error. With a large enough sample, t-ratios greater than 1.96 (in absolute value) suggest that your coefficient is statistically significantly different from 0 at the 95% confidence level. A threshold of 1.645 is used for 90% confidence.

Step-by-step explanation:

Hope it help you.

Answer:

Area = 156 square cm

Step-by-step explanation:

Answer:

a)CI 95 %  =  ( 6.7063  ;  6.7593)  ( in hundredths of an inch)

b) CI 95 %  =  ( 0.05 <  σ  <  0.089 ) ( in hundredths of an inch)

Step-by-step explanation:

From the problem statement, we have a manufacturing  process and we we assume a normal distribution, from sample data:

x =  6.7328        and  s  =  0.0644

a) CI 95 %  =  (  x ±  t(c) * s/√n )

t(c)    df =  n  -1    df  =  25 - 1     df = 24

CI  =  95 %   α = 5 %     α  = 0.05     α /2  =  0.025

Then  from t-student table   t(c) = 2.060

s/√n  =  0.0644/ √ 25      s/√n = 0.01288

CI 95 %  =  (  x ±  t(c) * s/√n )  =  ( 6.7328  ± 2.060*0.01288)

CI 95 %  = ( 6.7328 ±  0.02653 )

CI 95 %  =  ( 6.7063  ;  6.7593)  ( in hundredths of an inch)

b) CI 95 % of the variance  is:

CI 95 %  =  (  ( n - 1 ) * s² / χ²₁ - α/₂ ≤  σ²  ≤   ( n - 1 )*s² / χ²α/₂ )

( n - 1 ) * s²   = ( 25 - 1 ) * (0.0644)² =  24* 0.00414

( n - 1 ) * s²   = 0.09936

And from  χ²  table we look for values of

χ² α/₂ ,df    df = 24   and  α/2 =  0.025

χ² (0.025,24)  = 12.40      and    χ²₁ - α/₂  =   χ² (0.975, 24)

χ² (0.975, 24) = 39.36

Then

CI 95 %  = ( 0.09936 / 39.36  ≤  σ²  ≤   0.09936 / 12.40)

CI 95 %  = ( 0.0025  ≤   σ²  ≤  0.0080)

Then for the standard deviation,  we take the square root of that interval

CI 95 %  =  ( 0.05 <  σ  <  0.089 )

9514 1404 393

Answer:

 ∛188 ≈ 5.72865

Step-by-step explanation:

Any scientific or graphing calculator or spreadsheet can tell you the cube root of 188.

  ∛188 ≈ 5.72865431598...

__

You know that 5³ = 125 and 6³ = 216, so the root will lie between 5 and 6, closer to 6. As a first approximation, you can figure it is about ...

  x = ∛188 ≈ 5 + (188-5³)/(6³ -5³) = 5 + 63/89 ≈ 5.71

You can figure this much using a 4-function calculator.

A closer approximation (x') can be had using the iteration formula ...

  x' = (2x³ +188)/(3x²)

For x = 5.71, the value of x' is ...

  x' ≈ (2×5.71³ +188)/(3×5.71²) ≈ 560.3388/97.8123 ≈ 5.7287

This value is correct when the root is rounded to 4 decimal places. Another execution of the iteration formula using this value will give the root accurate to 9 decimal places.

Answer:

Answer in pic

Step-by-step explanation:

The required value of A₀ = 3 and k = ln 2, as of the given exponential model.

The function which is in format f(x) =a^x  where a is constant and x is variable,  the domain of this exponential function lies   (-∞, ∞).

here,
Since.
[tex]a^x = e^{xlna}[/tex]
According to the question,
[tex]3(2)^x = A_oe^{kx}\\3e^{xln2} = A_oe^{kx}[/tex]
Now

Compare the values,
A₀ = 3 and k = ln 2,

Thus, According to the stated exponential model, A0 = 3 and k = ln 2.

Learn more about exponential function here:

brainly.com/question/15352175

#SPJ2

Solution :

We have :

X            0     1          2    3   4

P(X)     0.10   A    0.35   B    C

a). P(X < 2) = 0.35

P(X < 2) = P(X = 0) + P(X = 1) = 0.35

⇒ 0.10 + A = 0.35

⇒ A = 0.25

So the value of A is 0.25

b). The total probability = 1

So ,

0.10 + A + 0.35 + B + C = 1

0.10 + 0.25 + 0.35 + B + C = 1

B + C  = 1 - 0.70

B + C  = 0.30  ......(i)

We have the expected value = 1.9

So, [tex]$\sum X P(X) = 19$[/tex]     for x  = 0, 1, 2, 3, 4

⇒ (0 x 0.10) + (1 x 0.25) + (2 x 0.35) + (3 x B) + (4 x C) = 1.9

⇒ 0 + 0.25 + 0.70 + 3B + 4C = 1.9

⇒ 3B + 4C = 1.9 - 0.95

⇒ 3B + 4C = 0.95   ...................(ii)

From (i), we take the value of B = 0.30 - C and substitute it in the equation (i), we get,

⇒ 3( 0.30 - C) + 4C = 0.95

⇒ 0.90 - 3C + 4C = 0.95

⇒ C = 0.95 - 0.90

       = 0.05

Now substituting the value of C = 0.05 in (ii), we get,

⇒ B = 0.05 = 0.30

⇒ B = 0.25

c). The value of C is 0.05

Answer:

the answer is 2

Step-by-step explanation: because 250 -22 is i dont even know

Answer:

55

Step-by-step explanation:

Usando el método de polya:

a. Que se pregunta?

b. ¿Cuáles son los datos dados?

c. ¿Cuál es la operación que se utilizará?

d. oración numérica:

e. Solución y respuesta completa:

[tex] \frac{3}{4} \div 2 \\ = \frac{3}{4} \times \frac{1}{2} \\ = \frac{3}{8} [/tex]

explicación paso a paso:

Forma decimal: 0.375

#CarryOnLearning

Answer:

50,000 millimeters

Step-by-step explanation:

Hope this helps : ) I have no explanation but it's correct

Answer:

c+12<16

c=16-12

c=4

answer is this

Given:

The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3.

The value of all the Rs5 coins is Rs45.

To find:

The value of all the Rs2 coins in the box.

Solution:

Let x be the number of Rs2 coins and y be the number of Rs5 coins.

The value of all the Rs5 coins is Rs45.

[tex]5y=45[/tex]

[tex]y=\dfrac{45}{5}[/tex]

[tex]y=9[/tex]

The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3.

[tex]\dfrac{x}{y}=\dfrac{1}{3}[/tex]

[tex]\dfrac{x}{9}=\dfrac{1}{3}[/tex]

Multiply both sides by 9.

[tex]\dfrac{x}{9}\times 9=\dfrac{1}{3}\times 9[/tex]

[tex]x=3[/tex]

The value of all the Rs2 coins in the box is:

[tex]\text{Required value}=2x[/tex]

[tex]\text{Required value}=2(3)[/tex]

[tex]\text{Required value}=6[/tex]

Therefore, the value of all the Rs. 2 coins in the box is Rs. 6.

Answer:

D. 20%

Step-by-step explanation:

2 years = 24 months

400 × 24 = 9600

9600 - 8000 = 1600

1600/8000 = 1/5

1/5 = 20%

9514 1404 393

Answer:

  a4 = -666

Step-by-step explanation:

Use the recursive definition repeatedly.

  a1 = 6

  a2 = -5(6) +4 = -26

  a3 = -5(-26) +4 = 134

  a4 = -5(134) +4 = -666