find out the Range coefficient of the range

marks number of students
20-29 8
30-39 12
40-49 20
50-59 7
60-69 3​

Answer:

9a²b

Step-by-step explanation:

Hi there!

We need to find the greatest common factor out of 45a²b³ and 18[tex]a^{4}[/tex]b

We can split apart the monomials to make it easier

45a²b³ is 45*a²b³

18[tex]a^{4}[/tex]b is 18*[tex]a^{4}[/tex]b

First, let's find the GCF out of 45 and 18 (the number coefficients)

we can find all of the multiples of the 2 numbers:

45 is made up of 9 and 5

9 is made up of 3 and 3

so 3*3*5 is 45

18 is made up of 2 and 9

9 is made up of 3 and 3

so 2*3*3 is 18

3*3 is in both 45 and 18, so 9 is the GCF out of 45 and 18

Now let's find the GCF out of a²b³ and [tex]a^{4}[/tex]b

a²b³ made up of a² and b³

so a²b³ is a*a*b*b*b

[tex]a^{4}[/tex]b is made up of [tex]a^{4}[/tex] and b

so [tex]a^{4}[/tex]b is a*a*a*a*b

a*a*b is in both a²b³ and [tex]a^{4}[/tex]b, so the GCF out of a²b³ and [tex]a^{4}[/tex]b is a²b

Now multiply 9 and a²b together, as they are only the GCF of the parts of the monomials

9*a²b=9a²b

there's the greatest common factor of the 2 monomials

Hope this helps!

Answer:

[tex]a_n = 4.5 * 3^{n-1}[/tex]

Step-by-step explanation:

Given

[tex]a_4 = 121.5[/tex]

[tex]r = 3[/tex]

Required

[tex]a_n = a_1 * r^{n -1}[/tex]

Substitute 4 for n in [tex]a_n = a_1 * r^{n -1}[/tex]

[tex]a_4 = a_1 * r^{4 -1}[/tex]

[tex]a_4 = a_1 * r^3[/tex]

Substitute 121.5 for [tex]a_4[/tex]

[tex]121.5 = a_1 * 3^3[/tex]

[tex]121.5 = a_1 * 27[/tex]

Solve for a1

[tex]a_1 = \frac{121.5}{27}[/tex]

[tex]a_1 = 4.5[/tex]

So, we have:

[tex]a_n = a_1 * r^{n -1}[/tex]

[tex]a_n = 4.5 * 3^{n-1}[/tex]

Answer:

First I substituted 121.5 for an, 4 for n, and 3 for r in the general form. Then I solved to find a1 = 4.5. Finally, I substituted 4.5 for a1 and 3 for r in the general form to get an = 4.5 ⋅ 3n−1.

Step-by-step explanation:

sample answer on edge ;)

Answer:

0.8743 = 87.43% probability that more than one accident occurs per year

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Buchtal, a manufacturer of ceramic tiles, reports on average 3.1 job-related accidents per year.

This means that [tex]\mu = 3.1[/tex]

What is the probability that more than one accident occurs per year?

This is:

[tex]P(X > 1) = 1 - P(X \leq 1)[/tex]

In which

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

Then

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3.6}*(3.6)^{0}}{(0)!} = 0.0273[/tex]

[tex]P(X = 1) = \frac{e^{-3.6}*(3.6)^{1}}{(1)!} = 0.0984[/tex]

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0273 + 0.0984 = 0.1257[/tex]

[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1257 = 0.8743[/tex]

0.8743 = 87.43% probability that more than one accident occurs per year

Answer:

59

Step-by-step explanation:

Just get the supplement of 121.

So angle 1 is 180-121 = 59 degrees

Answer:

y = -3x + 2

Step-by-step explanation:

If two lines are parallel to each other, they have the same slope.

The first line is y = -3x + 7. Its slope is -3. A line parallel to this one will also have a slope of -3.

Plug this value (-3) into your standard point-slope equation of y = mx + b.

y = -3x + b

To find b, we want to plug in a value that we know is on this line: in this case, it is (2, -4). Plug in the x and y values into the x and y of the standard equation.

-4 = -3(2) + b

To find b, multiply the slope and the input of x (2)

-4 = -6 + b

Now, add 6 to both sides to isolate b.

2 = b

Plug this into your standard equation.

y = -3x + 2

This equation is parallel/perpendicular to your given equation (y = -3x + 7) and contains point (2, -4)

Hope this helps!

(3.1) … … …

[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{2x-y}{x-2y}[/tex]

Multiply the right side by x/x :

[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{2-\dfrac yx}{1-\dfrac{2y}x}[/tex]

Substitute y(x) = x v(x), so that dy/dx = x dv/dx + v :

[tex]x\dfrac{\mathrm dv}{\mathrm dx} + v = \dfrac{2-v}{1-2v}[/tex]

This DE is now separable. With some simplification, you get

[tex]x\dfrac{\mathrm dv}{\mathrm dx} = \dfrac{2-2v+2v^2}{1-2v}[/tex]

[tex]\dfrac{1-2v}{2-2v+2v^2}\,\mathrm dv = \dfrac{\mathrm dx}x[/tex]

Now you're ready to integrate both sides (on the left, the denominator makes for a smooth substitution), which gives

[tex]-\dfrac12\ln\left|2v^2-2v+2\right| = \ln|x| + C[/tex]

Solve for v, then for y (or leave the solution in implicit form):

[tex]\ln\left|2v^2-2v+2\right| = -2\ln|x| + C[/tex]

[tex]\ln(2) + \ln\left|v^2-v+1\right| = \ln\left(\dfrac1{x^2}\right) + C[/tex]

[tex]\ln\left|v^2-v+1\right| = \ln\left(\dfrac1{x^2}\right) + C[/tex]

[tex]v^2-v+1 = e^{\ln\left(1/x^2\right)+C}[/tex]

[tex]v^2-v+1 = \dfrac C{x^2}[/tex]

[tex]\boxed{\left(\dfrac yx\right)^2 - \dfrac yx+1 = \dfrac C{x^2}}[/tex]

(3.2) … … …

[tex]y' + \dfrac yx = \dfrac{y^{-3/4}}{x^4}[/tex]

It may help to recognize this as a Bernoulli equation. Multiply both sides by [tex]y^{\frac34}[/tex] :

[tex]y^{3/4}y' + \dfrac{y^{7/4}}x = \dfrac1{x^4}[/tex]

Substitute [tex]z(x)=y(x)^{\frac74}[/tex], so that [tex]z' = \frac74 y^{3/4}y'[/tex]. Then you get a linear equation in z, which I write here in standard form:

[tex]\dfrac47 z' + \dfrac zx = \dfrac1{x^4} \implies z' + \dfrac7{4x}z=\dfrac7{4x^4}[/tex]

Multiply both sides by an integrating factor, [tex]x^{\frac74}[/tex], which gives

[tex]x^{7/4}z'+\dfrac74 x^{3/4}z = \dfrac74 x^{-9/4}[/tex]

and lets us condense the left side into the derivative of a product,

[tex]\left(x^{7/4}z\right)' = \dfrac74 x^{-9/4}[/tex]

Integrate both sides:

[tex]x^{7/4}z=\dfrac74\left(-\dfrac45\right) x^{-5/4}+C[/tex]

[tex]z=-\dfrac75 x^{-3} + Cx^{-7/4}[/tex]

Solve in terms of y :

[tex]y^{4/7}=-\dfrac7{5x^3} + \dfrac C{x^{7/4}}[/tex]

[tex]\boxed{y=\left(\dfrac C{x^{7/4}} - \dfrac7{5x^3}\right)^{7/4}}[/tex]

(3.3) … … …

[tex](\cos(x) - 2xy)\,\mathrm dx + \left(e^y-x^2\right)\,\mathrm dy = 0[/tex]

This DE is exact, since

[tex]\dfrac{\partial(-2xy)}{\partial y} = -2x[/tex]

[tex]\dfrac{\partial\left(e^y-x^2\right)}{\partial x} = -2x[/tex]

are the same. Then the general solution is a function f(x, y) = C, such that

[tex]\dfrac{\partial f}{\partial x}=\cos(x)-2xy[/tex]

[tex]\dfrac{\partial f}{\partial y} = e^y-x^2[/tex]

Integrating both sides of the first equation with respect to x gives

[tex]f(x,y) = \sin(x) - x^2y + g(y)[/tex]

Differentiating this result with respect to y then gives

[tex]-x^2 + \dfrac{\mathrm dg}{\mathrm dy} = e^y - x^2[/tex]

[tex]\implies\dfrac{\mathrm dg}{\mathrm dy} = e^y \implies g(y) = e^y + C[/tex]

Then the general solution is

[tex]\sin(x) - x^2y + e^y = C[/tex]

Given that y (1) = 4, we find

[tex]C = \sin(1) - 4 + e^4[/tex]

so that the particular solution is

[tex]\boxed{\sin(x) - x^2y + e^y = \sin(1) - 4 + e^4}[/tex]

Answer:

4

Problem:

If m and n are positive integers and m^2 - n^2 = 9, which of the following could be the value of

n?

A) 1

B) 16

C) 9

D) 4

Step-by-step explanation:

One approach would be to plug in the choices and see.

If n=1, then we have m^2-1=9.

This would give m^2=10 after adding 1 on both sides. There is no integer m when squared would give us 10. ( Square root of 9 is a decimal )

If n=16, then we would have m^2-256=9.

This would give m^2=265 after adding 256 on both sides. There is no integer m when squared would give us 265. ( Square root of 265 is a decimal )

If n=9, then we would have m^2-81=9.

This would give m^2=90 after adding 81 on both sides. There is no integer m when squared would give us 90. ( Square root of 90 is a decimal )

If n=4, then we would have m^2-16=9.

This would give m^2=25 after adding 16 on both sides. There is an integer m when squared would give us 25. ( Square root of 25 is a 5)

Answer:

Step-by-step explanation:

this is the famous dirt formula,  :P  I made it up   :D

D=rt    ( notice it looks like   Dirt  , kinda,  but it also means it dirt simple )

D= distance

r = rate  ( think speed or how fast)

t = time  ( in what ever units of time you want to use, seconds, minutes, hours )

13:28 - 11:20 = 128 minutes ( b/c the question is asking in MPH convert to hours)   2.4666667 hours

210 miles =  r * 2.46666667

210 / 2.46666667 = r   ( in MPH) ( does anyone else find it odd that they are saying miles in London instead of kilometers? :/ )

85.135 MPH = rate

so no, not even close to 104 MPH  :/  

Answer:

Average speed is 98 mph

Step-by-step explanation:

[tex]\frac{distance (miles)}{time (hours)}[/tex] = speed [tex]\frac{mile}{hours}[/tex]  (miles per hour is a ratio)

The time is 2 hours and 8 minutes.

[tex]\frac{8}{60}[/tex] = .13333 ( 8 minutes / 60 minutes in a hour)

So time is 2.133333 hours .

Divide the distance 210 by the time 2.13333 and get the speed.

Its 98.437..

Round to 98 miles per hour.

Answer:

5√2

Step-by-step explanation:

Given √10*√5

Using surd, the expression can be evaluated further as :

√10*√5 = √50

√50 can be expressed as :

√50 = √25*2 = √25 * √2

√25 = 5

Hence,

√50 = √25 * √2 = 5√2

Hence, √10*√5 = 5√2

Answer:

[tex]3.5[/tex]

Step-by-step explanation:

The smallest side of a triangle is formed by the smallest angle in the triangle.

To find the side opposite (formed by) the 20 degree angle, we can use the Law of Cosines. The Law of Cosines states that for any triangle, [tex]c^2=a^2+b^2-ab\cos \gamma[/tex], where [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] are the three sides of the triangle and [tex]\gamma[/tex] is the angle opposite to [tex]c[/tex].

Let [tex]c[/tex] be the side opposite to the 20 degree angle.

Assign variables:

Substituting these variables, we get:

[tex]c^2=4^2+7^2-2(4)(7)\cos 20^{\circ},\\c^2=16+49-56\cos 20^{\circ},\\c^2=12.377213236,\\c=\sqrt{12.377213236}=3.51812638147\approx \boxed{3.5}[/tex]

Therefore, the shortest side of this triangle is 3.5.

Answer:

(a) Residents

(b) [tex]Median = 0.13[/tex]

(c)  [tex]\bar x = 0.14[/tex]

(d) Right skewed

Step-by-step explanation:

Given

The data of residents without health insurance

Solving (a): The variable

The variable is the residents

Solving (b): The median

First, we sort the data

[tex]Sorted: 0.09, 0.10, 0.11, 0.13, 0.13, 0.14, 0.16, 0.19, 0.25[/tex]

So, the median position is:

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

[tex]Median = \frac{9 + 1}{2}[/tex]

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

[tex]Median = 5th[/tex]

The 5th element of the dataset is: 0.13

So:

[tex]Median = 0.13[/tex]

Solving (c): The mean

This is calculated as:

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

[tex]\bar x = \frac{0.09+ 0.10+ 0.11+ 0.13+ 0.13+ 0.14+ 0.16+ 0.19+ 0.25}{9}[/tex]

[tex]\bar x = \frac{1.3}{9}[/tex]

[tex]\bar x = 0.14[/tex]

Solving (d): The shape of the distribution

In (b) and (c), we have:

[tex]Median = 0.13[/tex]

[tex]\bar x = 0.14[/tex]

By comparison, the mean is greater than the median.

Hence, the shape is: right skewed.

Answer:

Dear the answer is 100% D

Good luck

Answer:

2

Step-by-step explanation:

Answer:

Common Ratio=-3

Step-by-step explanation:

To find any common ratio in a sequence, always take the second number in the sequence and divide it by the first number. However, you must be careful because the common ratio should only be negative if the values in the sequence are alternating between negative and positive. Therefore, if the sequence of numbers is simply decreasing in value, this does not mean that the common ratio is negative. The common ratio would still be positive. If the sequence is decreasing in value, this means that the common ratio would be a fraction or a decimal less than one.

Answer:

x = 1.5

Step-by-step explanation:

First, calculate 6-9, which is -3.

Then we add 9 on both sides so that on the left, we only have 4x, and on the right, we have 6.

Then divide by 4 on both sides to get x = 1.5

Answer:

Step-by-step explanation:

Dd

Answer:

 B and D are polynomial

Step-by-step explanation:

An algebraic expression with non-zero coefficients and variables having non-negative integers as exponents is called a polynomial.

A)

If it is [tex]1 -\frac{5x^{2}}{x }=1-5x[/tex]   , then it is a polynomial.

But if it is [tex]\frac{1-5x^{2}}{x}[/tex] then it is not a polynomial

Answer:

perimeter=9cm

Area=2cm^2

step by step explanation:

Firstly solve the sides that don't have figures using trigonometry

#1..sin15°=opposite/hypotenuse

sin15=opposite/4

opposite=sin15×4

=1.035, round off to 1cm

Then find the value of the base

tan15=opposite/adjacent

tan15=1/adjacent

1=tan15adjascent

1/tan15=adjacent

adjacent or base=3.7 round off to 4cm

After finding these values find the perimeter

p=side+side+side

p=4cm+4cm+1cm

p=9cm

Find the area

1/2bh

1/2×4×1

A=2cm2

Answer:

90deg +20deg.

Step-by-step explanation:

Angle RST can be broken down into RSQ and QST, whose measures are 90 deg and 20 deg, respectively. So you just add up the two parts to get the whole.

Hope this helps!

Answer:

f(x+1) = -3/4 × f(x)

Step-by-step explanation:

first of all, the sign of the numbers in the sequence is alternating. so, there must be a "-" involved.

that eliminates the first and third answer options.

and the absolute values of the numbers in the sequence are going down. |f(x+1)| < |f(x)|

that eliminates the fourth answer option, as this says that

|f(x)| < |f(x+1)|. and that is the opposite of how the actual sequence behaves.

Answer:

what is the question

pls mark me as brainlist

Thank you for the points

Answer:

1.19

Step-by-step explanation:

1+0.02+0.05+0.12 = 1.19

Answer:

Probability of defective televisions : Now, If two televisions are randomly​ selected, then the probability that both televisions work. Hence, the probability that both televisions work is 0.5289 . Hence, the probability at least one of the two televisions does not​ work is 0.4711.

Answer:

Below.

Step-by-step explanation:

22-8 = 14x2 = 28.

Answer:

10 is the correct answer

Answer:

Go with the third option 10!

i hope this helped!

Answer:

[tex]-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}[/tex]

Step-by-step explanation:

There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of [tex]6\cos(105^{\circ}})[/tex] is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.

Solution 1 (Cosine Addition Identity):

Nonetheless, to find the exact value we must use trigonometry identities.

Identity used:

[tex]\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta[/tex]

Notice that [tex]45+60=105[/tex] and therefore we can easily solve this problem if we know values of [tex]\cos(45^{\circ})[/tex], [tex]\cos(60^{\circ})[/tex], [tex]\sin (45^{\circ})[/tex], and [tex]\sin(60^{\circ})[/tex], which is plausible as they are all key angles on the unit circle.

Recall from either memory or the unit circle that:

Therefore, we have:

[tex]\cos(105^{\circ})=\cos(45^{\circ}+60^{\circ}}),\\\cos(45^{\circ}+60^{\circ}})=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ},\\\cos(45^{\circ}+60^{\circ}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}[/tex]

Since we want the value of [tex]6\cos 105^{\circ}[/tex], simply multiply this by 6 to get your final answer:

[tex]6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}[/tex]

Solution 2 (Combination of trig. identities):

Although less plausible, you may have the following memorized:

[tex]\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]

If so, we can use the following trig. identity:

[tex]\cos(\theta)=\sin(90^{\circ}-\theta)[/tex] (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)

Therefore,

[tex]\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})[/tex]

Recall another trig. identity:

[tex]\sin(-\theta)=-\sin (\theta)[/tex] and therefore:

[tex]\sin (-15^{\circ})=-\sin (15^{\circ})[/tex]

Multiply by 6 to get:

[tex]6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}[/tex] (alternative final answer).

x − 2 out of 5n ( x −2 ) + 8 ( x − 2) . ( x − 2 ) ( 5 n + 8 )

I hope this is correct and helps!

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\large\textsf{5n(x - 2) + 8(x - 2) =}[/tex]

[tex]\large\text{DISTRIBUTE 5 and 8 WITHIN the PARENTHESES}[/tex]

[tex]\large\textsf{= 5n(x) + 5(-2) + 8(x) + 8(-2)}[/tex]

[tex]\large\textsf{= 5nx - 10n + 8x - 16}[/tex]

[tex]\boxed{\boxed{\huge\textsf{Answer: \bf (x - 2)(5n + 8)}}}\huge\checkmark[/tex]

[tex]\large\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]