1. There are 50 contestants signed up for a TV show. There are 36 more female contestants than male contestants. How many female contestants have signed up to compete? Show your solution and explain how you plan to explain this to your students.

Answer:

x ≥ -6

Step-by-step explanation:

x - 4 ≤ 3x + 8

Add 4 to both sides.

x ≤ 3x + 12

Subtract 3x from both sides.

-2x ≤ 12

Divide both sides by -2. Remember to change the direction of the inequality sign since you are dividing by a negative number.

x ≥ -6

Answer:

[tex]x \geqslant - 6[/tex]

Step-by-step explanation:

[tex]x - 4 \leqslant 3x + 8 = = > \\ - 12 \leqslant 2x = = = > \\ x \geqslant - 6[/tex]

The function is illustrated below based on the information.

When x <= 8000

The cost remains constant at 0.35 when x increases from 0 to 8000. The slope of the cost function over this part is 0

When 8000 < x <= 20000

The cost remains constant at 0.75 when x increases from 8000 to 20000 and the slope of the cost function over this part is 0.

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[tex]order \: does \: not \:matter \\ sample \: space = 15 \: letters \\ no \: repetition\\ P(A) = 15C5 = 3003 \: ways[/tex]

Answer:

x = 9

Step-by-step explanation:

4x + 7 = 43

4x = 36

x = 9

The ordered pairs of the equation y = 3^x are (0,1), (1,3) and (2,9)

The equation is given as:

y = 3^x

Let x = 0, 1 and 2

y = 3^0 = 1

y = 3^1 = 3

y = 3^2 = 9

So, we have the following ordered pairs (0,1), (1,3) and (2,9)

Hence, the ordered pairs of the equation y = 3^x are (0,1), (1,3) and (2,9)

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Answer:

(0,3)

(-1,1)

Step-by-step explanation:

Hello!

Since both equations are equal to y, we can set the right-hand side of the equations equal to each other.

X can either be 0 or -1. Remember, a solution to a system is not complete without a y-value. Plug in 0 and -1 for x in the first equation, and find the corresponding y-values.

[tex]y = 2(0) + 3\\y = 3[/tex]

And

[tex]y = 2(-1) + 3\\y = -2 + 3\\y = 1[/tex]

So the solutions to the system are (0,3) and (-1,1)

Answer:

D.  I, II, and III

Step-by-step explanation:

A discontinuous function is a function which is not continuous.  

If f(x) is not continuous at x = a, then f(x) is said to be discontinuous at this point.

To prove whether a function is discontinuous, find where it is undefined.

A rational function is undefined when the denominator is equal to zero.

Therefore, to find the values that make a rational function undefined, set the denominator to zero and solve.

Function I

Denominator:  x - 2

Set to zero:  x - 2 = 0

Solve:  x = 2

Therefore, this function is undefined when x = 2 and so the function is discontinuous.

Function II

Denominator:  4x²

Set to zero:  4x² = 0

Solve:  x = 0

Therefore, this function is undefined when x = 0 and so the function is discontinuous.

Function III

Denominator:  x² + 3x + 2

Set to zero:  x² + 3x + 2 = 0

Solve:  

⇒ x² + 3x + 2 = 0

⇒ x² + x + 2x + 2 = 0

⇒ x(x + 1) + 2(x + 1) = 0

⇒ (x + 2)(x + 1) = 0

⇒ x = -2, x = -1

Therefore, this function is undefined when x = -2 and x = -1, and so the function is discontinuous.

Therefore, all three given functions are discontinuous.

Using the z-distribution, it is found that a sample of 171 should be selected.

The confidence interval is:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

The margin of error is:

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

In which:

For this problem, the parameters are:

[tex]z = 1.96, \sigma = 40, M = 6[/tex]

Hence we solve for n to find the needed sample size.

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

[tex]6 = 1.96\frac{40}{\sqrt{n}}[/tex]

[tex]6\sqrt{n} = 40 \times 1.96[/tex]

[tex]\sqrt{n} = \frac{40 \times 1.96}{6}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2[/tex]

n = 170.7.

Rounding up, a sample of 171 should be selected.

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Answer:

5 points

Step-by-step explanation:

Subtracting their scores, we get (-3)-(-8)=5

Answer:

B.) (1, -5)

Given, equation:

[tex]\sf \dfrac{(x-1)^2}{25} -\dfrac{(y+5)^2}{49} =1[/tex]

Comparing it with standard formula of hyperbola:

[tex]\sf \dfrac{(x-h)^2}{a^2} -\dfrac{(y-k)^2}{b^2} =1[/tex]

where (h, k) is center

The given equation in standard form:

[tex]\sf \dfrac{(x-1)^2}{5^2} -\dfrac{(y-(-5))^2}{7^2} =1[/tex]

Determined:

(h, k) = (1, -5)

a = 5, b = 7

So, here (1, -5) is the center of the given hyperbola.

The difference between the distance travelled by any point on the record and it's label is; 27.57cm.

According to the task content, it follows from observation that the quantity required is the difference between the circumference in each case and can be evaluated as follows;

Difference = π(17.78 - 9)

Difference = (22/7) × 8.78

Difference = 27.57cm.

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Answer:

  (a)  Theorem 9

Step-by-step explanation:

Any of the given theorems can be used to prove lines are parallel. We need to find the one that is applicable to the given geometry.

The marked angles are between the parallel lines (interior) and on opposite sides of the transversal (alternate).

Theorem 9 applies to congruent alternate interior angles.

I know you said "without making any assumptions," but this one is pretty important. Assuming you mean [tex]\alpha,\beta[/tex] are shape/rate parameters (as opposed to shape/scale), the PDF of [tex]X[/tex] is

[tex]f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}[/tex]

if [tex]x>0[/tex], and 0 otherwise.

The MGF of [tex]X[/tex] is given by

[tex]\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx[/tex]

Note that the integral converges only when [tex]t<2[/tex].

Define

[tex]I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx[/tex]

Integrate by parts, with

[tex]u = x^n \implies du = nx^{n-1} \, dx[/tex]

[tex]dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}[/tex]

so that

[tex]\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}[/tex]

Note that

[tex]I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}[/tex]

By substitution, we have

[tex]I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}[/tex]

and so on, down to

[tex]I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}[/tex]

The integral of interest then evaluates to

[tex]\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}[/tex]

so the MGF is

[tex]\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}[/tex]

The first moment/expectation is given by the first derivative of [tex]M_X(t)[/tex] at [tex]t=0[/tex].

[tex]\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}[/tex]

Variance is defined by

[tex]\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2[/tex]

The second moment is given by the second derivative of the MGF at [tex]t=0[/tex].

[tex]\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18[/tex]

Then the variance is

[tex]\Bbb V[X] = 18 - 4^2 = \boxed{2}[/tex]

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

[tex]M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k[/tex]

where [tex]M_X^{(k)}(0)[/tex] is the [tex]k[/tex]-derivative of the MGF evaluated at [tex]t=0[/tex]. This is also the [tex]k[/tex]-th moment of [tex]X[/tex].

Recall that for [tex]|t|<1[/tex],

[tex]\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k[/tex]

By differentiating both sides 7 times, we get

[tex]\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k[/tex]

Then the [tex]k[/tex]-th moment of [tex]X[/tex] is

[tex]M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}[/tex]

and we obtain the same results as before,

[tex]\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4[/tex]

[tex]\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18[/tex]

and the same variance follows.

Answer:

√2² cannot be answer yesah

Answer:

4.13

Step-by-step explanation:

Using sine ratio where it states that sine of a measurement is equal to ratio of opposite of measurement to hypotenuse (slant side/side with slope)

From the question, our opposite is x, hypotenuse is given as 15 units and the measurement is 16°. Apply the formula:

[tex]\displaystyle{\sin 16^{\circ} = \dfrac{x}{15}}[/tex]

Move 15 to multiply sin(16°):

[tex]\displaystyle{15\sin 16^{\circ} = x}[/tex]

Go ahead and input the value in a calculator since it's not a special triangle - a measurement that can be found using hand method - after you find the value, round it to 2 decimal places:

[tex]\displaystyle{x = 4.13456033...}[/tex]

Therefore, after rounding to 2 decimal places, the answer is:

[tex]\displaystyle{x=4.13}[/tex]

Therefore, the value of x is 4.13 units.

Answer: 6 %

Step-by-step explanation: 6 % of 12.5 = 0.75. 12.5 plus 0.75 = 13.25

The equivalent expressions are 6y + 4 and 11y + 16

Expression (A)

We have:

2y + 4(y + 1)

Open the bracket

2y + 4y + 4

Evaluate the like terms

6y + 4

Expression (B)

We have:

9y + 6 + 2(5 + y)

Open the bracket

9y + 6 + 10 + 2y

Evaluate the like terms

11y + 16

Hence, the equivalent expressions are 6y + 4 and 11y + 16

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Answer:

IV 5

Step-by-step explanation:

you have to solve for x:

x-3<2

x-3+3<2+3

x=5

could you put the brainlyest Icon for me? I have 1500 points, but no brainlyest, so I can't move up the ranks.

Answer:

Hello,

Step-by-step explanation:

[tex]log(5-2x)=0\\\\\ \Longleftrightarrow\ 10^{log(5-2x)}=10^0\\\\\\\Longleftrightarrow\ 5-2x=1\\\\\Longleftrightarrow\ 2x=5-1\\\\\\\Longleftrightarrow\ x=\dfrac{4}{2}\\\\\\\Longleftrightarrow\ \boxed{x=2}\\[/tex]

Answer:

None of these is not true

The height of the tide after eight hours is 16 feet.

From the information given, the function is given as 5cos (π/4)t + 11.

In this case, the time is given as eight hours. This will be:

f(8) = 5 × cos (π/4) × 8 + 11

f(8) = 5 × cos(2π) + 11

= (5 × 0.999) + 11

= 16 feet

The height is 16 feet.

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The total cost to laminate the table top is $128.

A trapezoid is a convex quadrilateral. A trapezoid has at least on pair of parallel sides.

The first step is to determine the area of the table top.

Area of a trapezoid = 0.5  x (sum of the lengths of the parallel sides) x height

0.5 x (6 + 10) x 4 = 32 square foot

Now, multiply the area by the cost per square foot

32 x 4 = $128

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Answer:

you have to put the formula of perimeter

Answer:

12a. L+8m + 14m

12b. 10m+8m +14m

12c. 66

12d.420m

12e.252m

quick questions

volume= 288cm cubed

The correct trigonometry formula to solve for x in the right angled triangle is sin(62°) = 18 / x

An equation is an expression that shows the relationship between two or more variables and numbers.

Trigonometric ratio is used to show the relationship between the sides and angles of a right angled triangle.

Hence:

sin(62°) = 18 / x

The correct trigonometry formula to solve for x in the right angled triangle is sin(62°) = 18 / x

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Answer:

  (f∘g)(x) = 27x² -90x +77

Step-by-step explanation:

The composition of functions (f∘g)(x) is interpreted to mean f(g(x)). The value is found the way any function value is found. The argument is substituted for the variable in the function definition.

After substituting g(x) for the argument of f(x), the resulting expression can be simplified.

  [tex](f\circ g)(x) =f(g(x))=f(3x-5)\\\\=3(3x-5)^2+2=3(9x^2-30x+25)+2\\\\\boxed{(f\circ g)(x)=27x^2-90x+77}[/tex]

Answer:

51

Step-by-step explanation:

Let c be the number of child tickets sold.

Let a be the number of adult tickets sold.

From the information given from the question, we can deduce:

6.10c + 9.70a = 1074.20 - Equation 1

c + a = 146 - Equation 2

From here, we can use the substitution method in solving simultaneous equations to find a and c. We will use Equation 2 as the base.

c = 146 - a - Equation 2A

We will then substitute equation 2A into Equation 1 to find a.

6.10(146-a) + 9.70a = 1074.20

890.6 - 6.10a + 9.70a = 1074.20

890.6 + 3.6a = 1074.20

3.6a = 1074.20 - 890.6

3.6a = 183.6

a = [tex]\frac{183.6}{3.6}[/tex]

a = 51

Therefore we know that 51  adult tickets were sold that day.

After that you can substitute a into equation 2 to find c.

it would take Janis 14 minutes to complete a package alone

From the information given,

Let's say that

Carol working time = x = 30 minutes

Janis working time = y

x + y = 16 minutes

Let's make 'y' the subject

30 - 16 = y

y= 14 minutes

Thus, it would take Janis 14 minutes to complete a package alone

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