What is the standard form equation of the ellipse that has vertices (-6, 2) and (14, 2) and foci (-3,2) and (11, 2)?

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:

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:

The answer is J.

Step-by-step explanation:

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:

Step-by-step explanation:

1) P= Area of Circle/ Area of large rectangle

Area of the circle = pi·r² = pi·2²=4 pi ft.²

Area of large rectangle=  l·w -12·10 =120 ft.²

P = 4pi/120 rewrite 120 as 4·30

P= 4 pi/4*30 = pi/30 = 3.14/40 ≈ .1047 ≈10% (because .1047·100 =10.47≅10)

2) P = Area of smaller rectangle/ Area of large rectangle

Area of smaller rectangle = l·w = 2·4 =8 ft.²

Area of large rectangle=l·w = 12·10=120 ft²

P= 8/120 ≅ .0666≅ 7% (because .0666·100 =6.66≅7)

3) P= Not the circle or smaller rectangle/ Area of large rectangle

Not the circle or smaller rectangle area

= Area of large rectangle - Area of circle -Area of smaller rectangle

= 120 -4·pi -8 = 120 - (4· 3.14) -8 = 99.4362939 ft²

Area of large rectangle = l·w = 12·10 =120 ft²

P = 99.4362939 /120 ≅ .8286 ≅83% (because .8286·100 =82.86≅83)

9514 1404 393

Answer:

  x = y + 360n, for any positive integer n

Step-by-step explanation:

Since x is greater than y, something must be added to y to get x. Angles have the same co-terminal ray at multiples of 360°. Then the amount added to y must be some multiple of 360°:

  x = y + 360n . . . . . for positive integer n

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:

We do not reject the Null Hypothesis

Step-by-step explanation:

From the question we are told that:

Sample size [tex]n=23[/tex]

Population mean [tex]\mu=9.02cm^3[/tex]

Sample mean [tex]\=x=8.16[/tex]

Standard deviation [tex]\sigma=0.7cm^3[/tex]

Significance level [tex]\alpha =0.01[/tex]

Generally the Null and and alternative Hypothesis are as follows

 [tex]H_0:\mu=9.02cm^3[/tex]

 [tex]H_a:\mu<9.02cm^3[/tex]

Therefore t critical Value is

 [tex]t\ critical\ Value=(\alpha,df)[/tex]

 [tex]t\ critical\ Value=(0.01,22)[/tex]

Where

  [tex]df=n-1\\\\df=23-1=>22[/tex]

Therefore

From t Table

 [tex]t value=-2.8[/tex]

Generally the equation for Z Critical is mathematically given by

 [tex]t=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

 [tex]t=\frac{8.16-9.02}{\frac{0.7}{\sqrt{23} } }[/tex]

 [tex]t=-5.89[/tex]

Therefore

Since the t test statistics is greater than the Critical value

Hence,we do not reject the Null Hypothesis

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 )

Answer:

28 children and 7 adults

Equation 1: a + c = 35

Equation 2: 6a + c = 70

Step-by-step explanation:

If the total number of people at the movie was 35 people, one of the equations will be a + c = 35.

If $70 was collected in total, the other equation will be 6a + c = 70.

Now, solve this system of equations:

a + c = 35

6a + c = 70

Solve by elimination by multiplying the top equation by -1, then adding the equations together:

-a - c = -35

6a + c = 70

Add these together, and solve for a:

5a = 35

a = 7

Since there were 35 people in total, find how many children attended by subtracting 7 from 35:

35 - 7

= 28

So, there were 28 children and 7 adults.

The equations used were: a + c = 35 and 6a + c = 70

So, the correct answer is:

28 children and 7 adults

Equation 1: a + c = 35

Equation 2: 6a + c = 70

Answer:

c+12<16

c=16-12

c=4

answer is this

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:

where is dx

I cannot find dx

Answer:

275, 350, 450, 550, 675

Step-by-step explanation:

Arrange in order

275, 300, 450, 490, 675

range 275 to 675

median 450

mean 438

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

Raise 300 to 350

Raise 490 to 550

New set of prices

275, 350, 450, 550, 675

range 275 to 675     same

median 450     same

mean 460    increased

Answer:

One set of prices could be: {275,300,450,600,675}

Another set could be: {275,300,450,600,675}

There are many other solutions possible.

====================================================

Explanation:

A = {450, 275, 675, 490, 300}

B = {275, 300, 450, 490, 675}

Set A is the original set of values in the order they were given to you. Set B is the sorted version of set A from smallest to largest.

The mean is found by adding up the values and dividing by 5 (because there are five items in the set).

The mean is (275+300+450+490+675)/5 = 2190/5 = 438. The shopkeeper wants to increase the mean to something larger, but keep the median and range the same.

The median is the middle most number. In set B, we can see that is 450. So the median is 450. We want to keep the median the same at 450.

The range is the difference in min and max

range = max - min = 675-275 = 400

We want to keep the range at 400

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

There are a number of ways to increase the mean, while keeping the median and range the same.

Let's say we keep the min and max the same. In order to increase the mean, we need to increase the 490 (second largest value) to something larger. Let's bump that up to 600 for instance.

Recomputing the mean gets us

(275+300+450+600+675)/5 = 2300/5 = 460

The old mean was 438 and the new mean is now 460. The mean has increased. This is due to the larger price pulling on the mean to get the mean to increase.

The median is still 450 because it's still in the direct middle of set C

C = {275,300,450,600,675}

The range is still the same as well because we haven't changed the min and max.

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

So one possible set could be

C = {275,300,450,600,675}

We could also have

D = {275,400,450,500,675}

The difference is that the 300 bumped to 400, and the 600 dropped to 500. You should find that the median and range are the same, while the mean is 460.

There are many possible solutions here.

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:

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

Step-by-step explanation:

Answer:

z (max)  =  34500 $

x₁  =  2

x₂  = 3

Step-by-step explanation:

Hilltop College  

3 hours per week  preparing lessons and grading papers

Serra College

4  hours per week  preparing lessons and grading papers

Total hours to spend per week preparing lessons  18

Let´s call  x₁  numbers of class at Hilltop College

and  x₂ numbers of class at Serra College then:

Objective function

z  =  6000*x₁  +  7500*x₂

Constraints:

1.-     x₁  +  x₂  ≤ 5        the total number of class

2.-    3*x₁  +  4*x₂  ≤  18

3. General constraints   x₁  ≥ 0       x₂    ≥  0   integers

After 6 iteration optimal solution is:  From on-line solver

z (max)  =  34500 $

x₁  =  2

x₂  = 3

Answer:

the answer is 2

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

Answer:

55

Step-by-step explanation:

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

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

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

Hey buddy I am here to help!

Yes these r similar!

Hope this helps!

Plz mark me brainliest!

Answer: integer is a whole number

Eg

10/2 = 5/1 = 5 = integer

But 12/8 = 3/2 = 1.5 = not an integer

So really, slope = integer

Means rise greater than run

And run is a factor of rise

Or run = 1 will satisfy the above too :)

Answer:

C. The run part of the slope is going to be one

Step-by-step explanation:

.

Anwer:$285

Explaination: Division method

$1995.00÷7=$285

Answer:

6.27

Step-by-step explanation:

We are to obtain the standard deviation of the given values :

{24, 18, 31,25, 34}

The standard deviation = √(Σ(x - mean)²/ n)

The mean = (ΣX) /n

Using calculator to save computation time :

The standard deviation, s = 6.27 (2 decimal places)

Answer:

D. 20%

Step-by-step explanation:

2 years = 24 months

400 × 24 = 9600

9600 - 8000 = 1600

1600/8000 = 1/5

1/5 = 20%

Answer:

Step-by-step explanation:

LA EDUCACION ES IMPIRTANTE YA QUE PROMUEVE UN MEJOR DESARROLLO DE LOS NIÑOS,NIÑAS Y ADOLESCENTES QUE LOS HACE FOMENTAR UN VINCULO MUY ESPECIAL CON SUS MAESTROS,COMPAÑEROS QUE LOS HACE SENTIR QUE SON PARTE DE SU FAMILIA ADEMAS ELLOS PASAN LA MAYORIA DE TIEMPO EN LA ESCUELA QUE LOS HACE SENTIRSE MÁS CÓMODOS COMO SI FUERA SU PROPIO HOGAR

Answer:

4 peaches

Step-by-step explanation:

$20/5 = 4 peaches