find r and s on the triangle​

Answer:

P=2pi×r

P=2×12pi=24pi

24pi÷4=6pi

6pi÷2=3pi

p=2×6×pi

p=12pi

12pi÷2=6pi

permiter=3pi+6pi+12=40.27

that is for part a

A perfect correlation is denoted by:

A. +1.0 and -1.0

Given:

Total fish (Sea bass and mackerel) = [tex]7\dfrac{2}{5}[/tex] pounds

Sea bass = [tex]4\dfrac{5}{8}[/tex] pounds

He gave away [tex]1\dfrac{7}{8}[/tex] pounds of mackerel.

To find:

The remaining mackerel.

Solution:

We know that,

Mackerel = Total fish - Sea bass

[tex]\text{Mackerel}=7\dfrac{2}{5}-4\dfrac{5}{8}[/tex]

[tex]\text{Mackerel}=\dfrac{37}{5}-\dfrac{37}{8}[/tex]

[tex]\text{Mackerel}=\dfrac{296-185}{40}[/tex]

[tex]\text{Mackerel}=\dfrac{111}{40}[/tex]

He gave away [tex]1\dfrac{7}{8}[/tex] pounds of mackerel. So, the remaining mackerel is:

[tex]\text{Remaining Mackerel}=\dfrac{111}{40}-1\dfrac{7}{8}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{111}{40}-\dfrac{15}{8}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{111-75}{40}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{36}{40}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{9}{10}[/tex]

Therefore, the remaining Mackerel is [tex]\dfrac{9}{10}[/tex] pounds or 0.9 pounds.

Answer:

The amount of Mackerel left is 9/10.

Step-by-step explanation:

total fish = 7 2/5 pounds

sea bass = 4 5/8

The amount of mackerel =

[tex]7\frac{2}{5}-4\frac{5}{8}\\\\=\frac{37}{5}-\frac{37}{8}\\\\=\frac{296-185}{40}\\\\=2 \frac{31}{40}[/tex]

Mackerel left =

[tex]2 \frac{31}{40}-1\frac{7}{8}\\\\= \frac{111}{40}-\frac{15}{8}\\\\=\frac{111-75}{40}\\\\=\frac{36}{40}\\\\=\frac{9}{10}[/tex]

Answer:

D = 120 Degrees ,  E : x = 14 , F: <JHK = 21, G: Summplementary Angle is 96 Degrees

Step-by-step explanation:

4X+2X = 180

6X=180

X=30

<ABD = 4X = 4(30) = 120

Answer:

six in parenthesis n minus two

Step-by-step explanation:

2 ways could be:

A. N minus two then multiply by six

B. Six times parentheses n minus two end parentheses

Answer: 0.38

Step-by-step explanation:

Since the variable x is represented by a standard normal distribution, the probability of x > 0.3 will be calculated thus:

P(x > 0.3)

Then, we will use a standard normal table

P(z > 0.3)

= 1 - p(z < 0.3)

= 1 - 0.62

= 0.38

Therefore, p(x > 0.3) = 0.38

The probability of x > 0.3 is 0.38.

Answer:

$3

Step-by-step explanation:

7 rands for $1

7×3=21

21 rands for $3

Answer:

$3

Step-by-step explanation:

7 rand = 1 dollar

Divide both sides by 7 rand.

1 = (1 dollar)/(7 rand)

Notice that the fraction (1 dollar)/(7 rand) equals 1, so multiplying by it will not change the amount, just the units.

21 rand * (1 dollar)/(7 rand) =

= 21/7 dollar

= 3 dollar = $3

Answer:

the answer is 3.162

Step-by-step explanation:

Answer:

The administrator should sample 968 students.

Step-by-step explanation:

We have to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.88}{2} = 0.06[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a p-value of [tex]1 - 0.06 = 0.94[/tex], so Z = 1.555.

Now, find the margin of error M as such

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

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Standard deviation of 300.

This means that [tex]n = 300[/tex]

If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample?

This is n for which M = 15. So

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

[tex]15 = 1.555\frac{300}{\sqrt{n}}[/tex]

[tex]15\sqrt{n} = 300*1.555[/tex]

Dividing both sides by 15

[tex]\sqrt{n} = 20*1.555[/tex]

[tex](\sqrt{n})^2 = (20*1.555)^2[/tex]

[tex]n = 967.2[/tex]

Rounding up:

The administrator should sample 968 students.

Answer:

geometric

Step-by-step explanation:

Answer:

The right answer is:

(a) 0.1456

(b) 18.125, 69.1202, 8.3139

Step-by-step explanation:

Given:

N = 24

n = 5

r = 7

The improperly drilled gearboxes "X".

then,

⇒ [tex]P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}[/tex]

(a)

P (all gearboxes fit properly) = [tex]P(x=0)[/tex]

                                               = [tex]\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}[/tex]

                                               = [tex]0.1456[/tex]

(b)

According to the question,

[tex]X = 91+5[/tex]

Mean will be:

⇒ [tex]\mu = E(x)[/tex]

       [tex]=E(91+5)[/tex]

       [tex]=9E(1)+5[/tex]

       [tex]=9.\frac{nr}{N}+5[/tex]

       [tex]=9.\frac{5.7}{24} +5[/tex]

       [tex]=18.125[/tex]

Variance will be:

⇒ [tex]\sigma^2=Var(X)[/tex]

         [tex]=V(9Y+5)[/tex]

         [tex]=81.V(Y)[/tex]

         [tex]=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}[/tex]

         [tex]=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}[/tex]

         [tex]=69.1202[/tex]

Standard deviation will be:

⇒ [tex]\sigma = \sqrt{69.1202}[/tex]

       [tex]=8.3139[/tex]      

Answer:

False

Step-by-step explanation:

A composite figure would be any irregular shapes and can be made up of multiple shapes

Answer:

The graph of y = f(x) will shift down 11 units

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

The residual for the point line (10,80) is option A 75. at the given line which best fits y = 6.2x + 13.

The graph follows a straight line equation shows a straight line graph.

            slope(m)=tan∅=y axis/x axis.

Straight line graphs show a linear relationship between the x and y values.

solving the equation:-

y = 6.2x + 13

putting value of X = 10

Y = 6.2 * 10 + 13

Y = 62 + 13

Y = 75

Hence the line best fit = 75.

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Answer:

x=57/2

Step-by-step explanation:

2x+3=60

<=> 2x=57

=> x=57/2

Answer:

it should be 3

Step-by-step explanation:

I hope this help

Step-by-step explanation:

We'll find the distance using the all-famous "Distance Formula." You'll probably come across it quite a bit, so it's best to have it written down somewhere.

The Distance Formula: [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]

Our points are (8, -3) and (4, -7), so we'll plug in those numbers accordingly.

For reference:

x2 = 4

x1 = 8

y2 = -7

y1 = -3

The calculation:

(substitute)

[tex]\sqrt{(4-8)^2+((-7)-(-3))^2 }[/tex]

(simplify)

[tex]\sqrt{(-4)^2+(-4)^2 }[/tex]

(square things)

[tex]\sqrt{16+16 }[/tex]

(add)

[tex]\sqrt{32}[/tex]

Answer:

[tex]\sqrt{32}[/tex]

Answer:

[tex]\boxed {\boxed {\sf C. \sqrt{32}}}[/tex]

Step-by-step explanation:

The distance between 2 points can be determined with the following formula.

[tex]d= \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2[/tex]

In this formula, (x₁, y₁) and (x₂, y₂) are the 2 points. We want to find the distance between the points (8, -3) and (4, -7). If we match the value with its corresponding variable, then we see:

Substitute the values into the formula.

[tex]d= \sqrt{(4-8)^2+(-7--3)^2[/tex]

Solve inside the parentheses.

[tex]d= \sqrt {(-4)^2+(-4)^2[/tex]

Solve the exponents.

[tex]d= \sqrt {16+16[/tex]

Add.

[tex]d= \sqrt {32}[/tex]

This radical can be simplified, but since it is an answer choice, we can leave it as is.

The distance between the points (8, -3) and (4, -7) is √32 and choice C is correct.

Answer:

9.425 or 3π

Step-by-step explanation:

arc length = 2πr(θ/360)

The angle is 90˚ as indicated by the square symbol.

ABC = 2π6(90/360) = 9.425 or 3π

Answer:

x = 960

Step-by-step explanation:

x=√{576×(576+1024)}

or, x = √(576×1600)

or, x = √576×√1600

or, x = 24×40

or, x = 960

Answered by GAUTHMATH

Answer:

Step-by-step explanation:

Answer:

Answer:

The probability that the average score of the 49 golfers exceeded 62 is 0.3897

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Therefore, the number 6.4 rounded to the nearest whole number is 6. * If the number you are rounding off is followed by 0,1,2,3,4, round the number down. To find 6.4 rounded to the nearest whole number.

Please Mark me brainliest

Answer:

Poggers

Step-by-step explanation:

Answer:

C.

Step-by-step explanation:

Science questions are T/F. That's .5 or [tex]\frac{1}{2}[/tex] chance to get it right.

7 questions. Each with a [tex]\frac{1}{2}[/tex].

That's [tex](\frac{1}{2})^{7}[/tex]  

[tex](\frac{1}{2})^{7}[/tex] = [tex]\frac{1}{128}[/tex] = .0078 = .008 rounded

English questions are multiple choice. 4 choices. That's .25 or [tex]\frac{1}{4}[/tex]

3 questions. Each [tex]\frac{1}{4}[/tex]

That's [tex](\frac{1}{4})^{3}[/tex]

[tex](\frac{1}{4})^{3}[/tex] = [tex]\frac{1}{64}[/tex] = .0156 = .016 rounded

The probability of getting all questions correct is 0.004 which is lower than for science (0.016).

Science questions are said to have a 6 out of 6 correct answer range and a 0.5 chance of being answered properly.

So, P(6) = 0.5⁶

P(6) = 0.016

We are informed that there are 4 out of 4 correct answers for English, with a 0.25 chance of getting it right.

Thus; P(4) = 0.25⁴ = 0.004

Thus, we conclude that they will have to study English.

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Answer:

Kindly check explanation

Step-by-step explanation:

The data table :

Use of SM _UK _China _ Russia _US _ col total

Yes ______480 __215 ___343 __640 _ 1678

No _______320 _ 285 ___357__ 360 _ 1322

Row Total__ 800 _500 __ 700_ 1000_ 3000

H0 : p1 = p2 = p3 = p4

H1 : p1 ≠ p2 ≠ p3 ≠ p4

Test statistic :

χ² = Σ(observed - Expected)² / Expected

Expected value of each cell = (Row total * column total) / N

N = grand total

Expected Values:

447.467 _279.667 _ 391.533 _ 559.333

352.533 _ 220.333 _ 308.467 _ 440.667

χ²=(2.36536+14.9527+6.01605+11.6337+3.00232 18.9793+7.63611+14.7665) = 79.352

Degree of freedom, df = (row-1)*(column-1) = (2 - 1) * (4 - 1) = 1 * 3 = 3

Using the Pvalue from Chisquare calculator :

Pvalue(79.352, 3) = 0.00000000001

Decision region :

Reject H0 ; If Pvalue < α

α = 0.05

Since 0.000000001 < 0.05 ; Reject H0 and conclude that not all population proportion are equal.

Sample proportion :

Phat = number of yes, x / total

For UK, Phat = 480/800 = 0.6

For China , Phat = 215/500 = 0.43

For Russia , Phat = 343/700 = 0.49

For US, Phat = 640/1000 = 0.64

Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this question:

Fleet of 18 means that [tex]N = 18[/tex]

9 are carrying nuclear weapons, which means that [tex]k = 9[/tex]

9 are destroyed, which means that [tex]n = 9[/tex]

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]

In which

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021[/tex]

[tex]P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666[/tex]

Then

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687[/tex]

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Answer:

The third side is increasing at an approximate rate of about 0.444 meters per minute.

Step-by-step explanation:

We are given a triangle with two sides having constant lengths of 13 m and 19 m. The angle between them is increasing at a rate of 2° per minute and we want to find the rate at which the third side of the triangle is increasing when the angle is 60°.

Let the angle between the two given sides be θ and let the third side be c.

Essentially, given dθ/dt = 2°/min and θ = 60°, we want to find dc/dt.

First, convert the degrees into radians:

[tex]\displaystyle 2^\circ \cdot \frac{\pi \text{ rad}}{180^\circ} = \frac{\pi}{90}\text{ rad}[/tex]

Hence, dθ/dt = π/90.

From the Law of Cosines:

[tex]\displaystyle c^2 = a^2 + b^2 - 2ab\cos \theta[/tex]

Since a = 13 and b = 19:

[tex]\displaystyle c^2 = (13)^2 + (19)^2 - 2(13)(19)\cos \theta[/tex]

Simplify:

[tex]\displaystyle c^2 = 530 - 494\cos \theta[/tex]

Take the derivative of both sides with respect to t:

[tex]\displaystyle \frac{d}{dt}\left[c^2\right] = \frac{d}{dt}\left[ 530 - 494\cos \theta\right][/tex]

Implicitly differentiate:

[tex]\displaystyle 2c\frac{dc}{dt} = 494\sin\theta \frac{d\theta}{dt}[/tex]

We want to find dc/dt given that dθ/dt = π/90 and when θ = 60° or π/3. First, find c:

[tex]\displaystyle \begin{aligned} c &= \sqrt{530 - 494\cos \theta}\\ \\ &=\sqrt{530 -494\cos \frac{\pi}{3} \\ \\ &= \sqrt{530 - 494\left(\frac{1}{2}\right)} \\ \\&= \sqrt{283\end{aligned}[/tex]

Substitute:

[tex]\displaystyle 2\left(\sqrt{283}\right) \frac{dc}{dt} = 494\sin\left(\frac{\pi}{3}\right)\left(\frac{\pi}{90}\right)[/tex]

Solve for dc/dt:

[tex]\displaystyle \frac{dc}{dt} = \frac{494\sin \dfrac{\pi}{3} \cdot \dfrac{\pi}{90}}{2\sqrt{283}}[/tex]

Evaluate. Hence:

[tex]\displaystyle \begin{aligned} \frac{dc}{dt} &= \frac{494\left(\dfrac{\sqrt{3}}{2} \right)\cdot \dfrac{\pi}{90}}{2\sqrt{283}}\\ \\ &= \frac{\dfrac{247\sqrt{3}\pi}{90}}{2\sqrt{283}}\\ \\ &= \frac{247\sqrt{3}\pi}{180\sqrt{283}} \\ \\ &\approx 0.444\text{ m/min}\end{aligned}[/tex]

The third side is increasing at an approximate rate of about 0.444 meters per minute.

9514 1404 393

Answer:

  0.444 m/min

Step-by-step explanation:

I find this kind of question to be answered easily by a graphing calculator.

The length of the third side can be found using the law of cosines. If the angle of interest is C, the two given sides 'a' and 'b', then the third side is ...

  c = √(a² +b² -2ab·cos(C))

Since C is a function of time, its value in degrees can be written ...

  C = 60° +2t° . . . . . where t is in minutes, and t=0 is the time of interest

Using a=13, and b=19, the length of the third side is ...

  c(t) = √(13² +19² -2·13·19·cos(60° +2t°))

Most graphing calculators are able to compute a numerical value of the derivative of a function. Here, we use the Desmos calculator for that. (Angles are set to degrees.) It tells us the rate of change of side 'c' is ...

  0.443855627418 m/min ≈ 0.444 m/min

_____

Additional comment

At that time, the length of the third side is about 16.823 m.

__

c(t) reduces to √(530 -494cos(π/90·t +π/3))

Then the derivative is ...

  [tex]c'(t)=\dfrac{494\sin{\left(\dfrac{\pi}{90}t+\dfrac{\pi}{3}\right)}\cdot\dfrac{\pi}{90}}{2\sqrt{530-494\cos{\left(\dfrac{\pi}{90}t+\dfrac{\pi}{3}}}\right)}}}\\\\c'(0)=\dfrac{247\pi\sqrt{3}}{180\sqrt{283}}\approx0.443855...\ \text{m/min}[/tex]

Answer:

a) someone will steal her stuff in any other situation even if a stranger watches it

Given:

The five number summary of two data sets are given as:

a) 0, 4, 12, 14, 20

b) 2, 8, 14, 18, 20

To find:

The range for the outliers.

Solution:

We know that,

An observation is considered an outlier if it is below [tex]Q_1-1.5(IQR)[/tex]

An observation is considered an outlier if it is above [tex]Q_3+1.5(IQR)[/tex]

Where, IQR is the interquartile range and [tex]IQR=Q_3-Q_1[/tex].

The five number summary of two data sets are given as:

0, 4, 12, 14, 20

Here, [tex]Q_1=4[/tex] and [tex]Q_3=14[/tex].

Now,

[tex]IQR=14-4[/tex]

[tex]IQR=10[/tex]

The range for the outliers is:

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[4-1.5(10),14+1.5(10)][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[4-15,14+15][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[-11,29][/tex]

An observation is considered an outlier if it is below -11.

An observation is considered an outlier if it is above 29.

The five number summary of two data sets are given as:

2, 8, 14, 18, 20

Here, [tex]Q_1=8[/tex] and [tex]Q_3=18[/tex].

Now,

[tex]IQR=18-8[/tex]

[tex]IQR=10[/tex]

The range for the outliers is:

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[8-1.5(10),18+1.5(10)][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[8-15,18+15][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[-7,33][/tex]

An observation is considered an outlier if it is below -7.

An observation is considered an outlier if it is above 33.