In a high school graduating class of 300, 200 students are going to college, 40 are planning to work full-time, and 80 are taking a gap year.

a. These are mutually exclusive events.
b. These are not mutually exclusive events.
c. You should add their individual probabilities.
d. None of the above are true.

Answer:

divide 99 by 5

99/5= 19.8

9514 1404 393

Answer:

  x = 2 or x = 9

Step-by-step explanation:

To use the quadratic formula, we first need the equation in standard form for a quadratic. We can get there by multiplying the equation by 3(x -3).

  2(3) +4(3(x -3)) = (x +4)(x -3)

  6 +12x -36 = x² +x -12

  x² -11x +18 = 0

Using the quadratic formula to find the solutions, we have ...

  [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-11)\pm\sqrt{(-11)^2-4(1)(18)}}{2(1)}\\\\x=\dfrac{11\pm\sqrt{49}}{2}=\{2,9\}[/tex]

The solutions are x=2 and x=9.

30.8 m

Step-by-step explanation:

Given: [tex]V = 43\:\text{m}[/tex], [tex]V_y = 30\:\text{m}[/tex]

The x-component of vector [tex]\vec{\text{V}}[/tex] is

[tex]V_x = \sqrt{V^2 - V_y^2} = \sqrt{(43)^2 - (30)^2} = 30.8\:\text{m}[/tex]

The x- component of the vector is 30.8meters.

If [tex]v = < a. b >[/tex] be a position vector then the magnitude of vector v is found by  [tex]|v| =\sqrt{a^{2}+b^{2} } }[/tex] , where a and b are the x and y component respectively.

And the direction is equals to the angle formed x- axis or y axis.

According to the given question

We have

Magnitude of the vector, |v| = 43meters

Y- component of the vector, b = 30meters

Since, we know that

[tex]|v| =\sqrt{a^{2} +b^{2} }[/tex]

Substitute the value of |v| = 43 and b = 30 in the above formula of magnitude.

⇒ [tex]43 = \sqrt{a^{2}+30^{2} }[/tex]

⇒ [tex]43 = \sqrt{a^{2}+900 }[/tex]

⇒ [tex]43^{2} =a^{2} + 900[/tex]

⇒ [tex]1849 = a^{2} + 900[/tex]

⇒ [tex]1849-900=a^{2}[/tex]

⇒ [tex]949=a^{2}[/tex]

⇒ [tex]a =\sqrt{949}[/tex]

⇒ [tex]a = 30.8[/tex]

Hence, the x- component of the vector is 30.8meters.

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You're looking for a solution of the form

[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n[/tex]

Differentiating twice yields

[tex]\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/tex]

[tex]\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n[/tex]

Substitute these series into the DE:

[tex]\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0[/tex]

[tex]\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

[tex]\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0[/tex]

which indicates that the coefficients in the series solution are governed by the recurrence,

[tex]\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}[/tex]

Use the recurrence to get the first few coefficients:

[tex]\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}[/tex]

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots[/tex]

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

[tex]\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}[/tex]

Just to confirm this solution is valid: we have

y = 10x - 7eˣ   ==>   y (0) = 0 - 7 = -7

y' = 10 - 7eˣ   ==>   y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.

9514 1404 393

Answer:

  t = (V -6(3.14)c^2)/(4(3.14)c)

Step-by-step explanation:

Isolate the term containing t, then divide by the coefficient of t

  [tex]V=4(3.14)ct+6(3.14)c^2\\\\V-6(3.14)c^2=t(4(3.14)c)\\\\\boxed{t=\dfrac{V-6(3.14)c^2}{4(3.14)c}}[/tex]

Answer:

100 different shoes

Step-by-step explanation:

4 styles * 5 sizes * 5 colors

4*5*5 = 100

Answer:

77/306 or around 25.2%

Step-by-step explanation:

[tex]\frac{7}{18} *\frac{11}{17}[/tex] section 1/total * section 2/(total-1) since there is no replacement

just solve and you get 77/306

Answer:

B

Step-by-step explanation:

For l1 and l2 to be parallel, these two angles need to be equal. 3x-15=2x+10, x=25

Answer:

14m

Step-by-step explanation:

[tex]7m-7+7m-7\\[/tex]

First, we need to eliminate the like term and collect the like term.

[tex]-7+-7=0[/tex]

Now, we have 7m +7m, sum them up and you will get the answer.[tex]7m+7m=14m[/tex]

So, the answer is 14m.

Answer:

14m

Step-by-step explanation:

7m -7 +7m +7​

7m + 7m - 7 + 7

14m

Answer:

K(t)=37

Step-by-step explanation:

k(t) = 13t - 2

k(3) = 13(3) - 2

k(3) = 39 - 2

k(3) = 37 <===

Answer:

48

Step-by-step explanation:

Pythagorean Theorem = h^2=p^2+b^2

We have,

(Hypotenuse)h=50

Let 14 be p, i.e (Perpendicular ,Known side)p=14

(Remaining side ,base)b=?(

Now,  

h^2=p^2+b^2

or, 50^2=14^2+b^2

or, 2500-196=b^2

or, √2304=b

b=48

Using the fundamental counting principle, it is found that: 50 2-scoop combinations are possible.

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

The flavors and the toppings are independent, which means that the fundamental counting principle is used to solve this question, which states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are p*q ways to do both things.

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

In this question:

So,

[tex]10 \times 5 = 50[/tex]

50 2-scoop combinations are possible.

A similar question is found at https://brainly.com/question/24067651

Answer:

x = -11/5 or -2.2

y = 13/5 or 2.6

Step-by-step explanation:

well, start by doing the multiplication. then we will see better.

2x + 2yi + xi + yii = -7 + 3i

2x + 2yi + xi - y = -7 + 3i

this is because, remember, i = sqrt(-1), and ii = -1.

now we group the i-factors and the terms without i and compare it to the corresponding parts on the right side.

2x - y = -7

2yi + xi = 3i

=> 2y + x = 3

x = 3 - 2y

and that we use ihr the first equation again

2×(3-2y) - y = -7

6 - 4y - y = -7

-5y = -13

y = 13/5

x = 3 - 2×13/5 = 3 - 26/5 = 15/5 - 26/5 = -11/5

Answer:

a. 3/4 of pocket money left

b. R10

Step-by-step explanation:

a. 4/4 - 1/4 = 3/4

b. 40/4 = 10 = 1/4 of pocket money

(a). The fraction of pocket money left is 3/4

(b). If he had R40 pocket money, he spend R10.

The ratio is defined as a relationship between two quantities, it is expressed one divided by the other.

Arithmetic operations can also be specified by the subtract, divide, and multiply built-in functions.

The operator that perform arithmetic operation are called arithmetic operators .

Operators which let do basic mathematical calculation

+ Addition operation : Adds values on either side of the operator.

For example 4 + 2 = 6

- Subtraction operation : Subtracts right hand operand from left hand operand.

for example 4 -2 = 2

* Multiplication operation : Multiplies values on either side of the operator

For example 4*2 = 8

Mandla spent one-quarter of his pocket money on sweets

Let x be the total amount of pocket money Mandla had originally

Solution of (a).

⇒ x - (1/4)x

⇒ (3/4)x

The fraction of pocket money left = 3/4

Solution of (b).

1/4 of pocket money

⇒ (1/4)x

⇒ (1/4)40

⇒ 10

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The expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters

From the question, the side lengths are

(4.6x-4.4y)(4.6x−4.4y) cm, (7.5x-8.8z)(7.5x−8.8z) cm, and  (7.7z-9.2y)(7.7z−9.2y) cm.

First, we will clear the brackets one after the other

For  (4.6x-4.4y)(4.6x−4.4y) cm

[tex]4.6x(4.6x-4.4y) -4.4y(4.6x-4.4y)[/tex]

[tex]21.16x^{2} -20.24xy -20.24xy+19.36y^{2}[/tex]

[tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex]

∴ (4.6x-4.4y)(4.6x−4.4y) cm = [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm

For  (7.5x-8.8z)(7.5x−8.8z) cm

[tex]7.5x(7.5x-8.8z) -8.8z(7.5x-8.8z)[/tex]

[tex]56.25x^{2} - 66xz -66xz + 77.44z^{2}[/tex]

[tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]

∴ (7.5x-8.8z)(7.5x−8.8z) cm = [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]  cm

For (7.7z-9.2y)(7.7z−9.2y) cm

[tex]7.7z(7.7z-9.2y)-9.2y(7.7z-9.2y)[/tex]

[tex]59.29z^{2} - 70.84yz-70.84yz+84.64y^{2}[/tex]

[tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex]

∴ (7.7z-9.2y)(7.7z−9.2y) cm =  [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm

Now, for the expression that represents the perimeter of the triangle,

Perimeter of a triangle can be calculated by determining the sum of all its sides

That is,

Perimeter of the triangle =  [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm + [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]  cm + [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm

[tex]=21.16x^{2} -40.48xy+19.36y^{2} + 56.25x^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz+84.64y^{2}[/tex]

Collect like terms

[tex]= 21.16x^{2}+ 56.25x^{2} -40.48xy+19.36y^{2}+84.64y^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz[/tex]

[tex]= 77.41x^{2} -40.48xy+104y^{2} - 132xz + 136.73z^{2} - 141.64yz[/tex]

[tex]= 77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm

Hence, the expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters

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9514 1404 393

Answer:

  the angle is 30°

Step-by-step explanation:

The area of a triangle given two sides and the included angle is ...

  A = 1/2ab·sin(C)

  5 cm² = 1/2(4 cm)(5 cm)sin(θ)

  0.5 = sin(θ) . . . . . . divide by 10 cm²

  θ = arcsin(.5) ≈ 30°

_____

Additional comment

An obtuse angle of 150° in that location will give a triangle with the same area.

Answer:

83

Step-by-step explanation:

1∇2= (2+2^1)

=2+2=4

(4)∇3= (2+3^4)

=2+81

=83

Answer:

0.40905 - 0.10204 = .30701 = 30.7 %

Step-by-step explanation:

0.23 0.40905

1.27 0.10204

9514 1404 393

Explanation:

You can check your answer by making sure that each of the primes you found is actually a prime. (Compare to a list of known primes, for example.) After you have determined your factors are primes, multiply them together to see if the result is 73. If so, you have found the correct prime factorization.

__

Additional comment

73 is prime, so its prime factor is 73.

  73 = 73

It would take 17 months and 14 days for the investment to grow to $1545.

To determine the time in months it would take for a $ 1500 dollar investment in a TFSA to grow to $ 1545 if the simple interest at a rate paid was 2% per annum, the following calculation must be performed:

First, you must obtain 2% of 1545 to determine the interest generated per year.

Then, a cross multiplication must be carried out considering the number of months it took to generate said interest, and compare it with the interest that arises from the subtraction of 1545 - 1500, that is, 45.

Therefore, it would take 17 months and 14 days for a $ 1500 dollar investment in a TFSA to grow to $ 1545.

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Answer:

[tex]V = \frac{4}{3} \pi {r}^{3} \\ \\ 3V = 4\pi {r}^{3} \\ {r}^{3} = \frac{3V}{4\pi} \\ \\ r = \sqrt[3]{ \frac{3V}{4\pi} } [/tex]

9514 1404 393

Answer:

  9 square inches

Step-by-step explanation:

The area is proportional to the square of the linear scale factor. We can use this to write the proportion ...

  A/(144 m²) = ((2 in)/(8 m))²

  A = (144·4/64) in² = 9 in²

The area representing 144 square meters is 9 square inches.

Answer:

Step-by-step explanation: Hello! Do

Answer:

[tex] \rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} [/tex]

Step-by-step explanation:

we would like to figure out the differential coefficient of [tex]e^{2x}(1+\ln(x))[/tex]

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

[tex] \displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )[/tex]

to do so distribute:

[tex] \displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x} [/tex]

take derivative in both sides which yields:

[tex] \displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )[/tex]

by sum derivation rule we acquire:

[tex] \rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x} [/tex]

Part-A: differentiating $e^{2x}$

[tex] \displaystyle \frac{d}{dx} {e}^{2x} [/tex]

the rule of composite function derivation is given by:

[tex] \rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)[/tex]

so let g(x) [2x] be u and transform it:

[tex] \displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x[/tex]

differentiate:

[tex] \displaystyle {e}^{u} \cdot 2[/tex]

substitute back:

[tex] \displaystyle \boxed{2{e}^{2x} }[/tex]

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

[tex] \displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)[/tex]

let

substitute

[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x} [/tex]

differentiate:

[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }[/tex]

Final part:

substitute what we got:

[tex] \rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }[/tex]

and we're done!

Answer:

Product Rule for Differentiation

[tex]\textsf{If }y=uv[/tex]

[tex]\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}[/tex]

Given equation:

[tex]y=e^{2x}(1+\ln x)[/tex]

Define the variables:

[tex]\textsf{Let }u=e^{2x} \implies \dfrac{du}{dx}=2e^{2x}[/tex]

[tex]\textsf{Let }v=1+\ln x \implies \dfrac{dv}{dx}=\dfrac{1}{x}[/tex]

Therefore:

[tex]\begin{aligned}\dfrac{dy}{dx} & =u\dfrac{dv}{dx}+v\dfrac{du}{dx}\\\\\implies \dfrac{dy}{dx} & =e^{2x} \cdot \dfrac{1}{x}+(1+\ln x) \cdot 2e^{2x}\\\\& = \dfrac{e^{2x}}{x}+2e^{2x}(1+\ln x)\\\\ & = \dfrac{e^{2x}}{x}+2e^{2x}+2e^{2x} \ln x\\\\& = e^{2x}\left(\dfrac{1}{x}+2+2 \ln x \right)\end{aligned}[/tex]

Answer:

16

Step-by-step explanation:

Inverse variation is of the form

xy = k where k is a constant

x=4 and y = 32

4*32 = k

128 = k

xy = 128

Let x = 8

8y = 128

Divide each side by 8

8y/8 = 128/8

y =16

Answer:  Choice C

This is because the input n = 2 leads to the output n-2 = 2-2 = 0

As another example: the input n = 4 leads to the output n-2 = 4-2 = 2

Whatever the input is, subtract 2 from it to get the output.

with no further informations, just go by looking at it.

it's 90°, all other options are too far off

Answer:

Spherical-Like Shape

Step-by-step explanation:

An s-orbital is spherical with the nucleus at its center.

Answer:

2 is the factor of every even number hope this help you