Meghan sells advertisements for a radio station. Each 30 second ad costs $20 per play, and each 60 second ad
costs $35 per play. Meghan sold 12 ads for $315. She wrote the system below letting x represent the number of 30
second ads and y represent the number of 60 second ads.
X+ y = 12
20x+35y = 315
What is the solution to the system of equations?

Need answers ASAP!!!!

Answer:

the answer of r is 8 i hope it will help

Answer:

hey hi mate

hope you like it

plz mark it as brainliest

Answer:

AB = 5.6 cm

Step-by-step explanation:

Reference angle (θ) = 62°

Hypotenuse = 12 cm

Adjacent = AB

Apply the trigonometric ratio formula, CAH, which is:

Cos θ = Adj/Hyp

Plug in the values

Cos 62° = AB/12

12*Cos 62° = AB

5.63365876 = AB

AB = 5.6 cm (1 decimal place)

Answer:

11 hours is right answer i hope it will help you

You have some data points labeled by [tex]x[/tex]. They form the set {3, 5, 7}.

The mean, [tex]\bar x[/tex], is the average of these values:

[tex]\bar x = \dfrac{3+5+7}3 = \dfrac{15}3 = 5[/tex]

Then in the column labeled [tex]x-\bar x[/tex], what you're doing is computing the difference between each data point [tex]x[/tex] and the mean [tex]\bar x[/tex]:

[tex]x=3 \implies x-\bar x = 3 - 5 = -2[/tex]

[tex]x=5 \implies x-\bar x = 5-5 = 0[/tex]

[tex]x=7 \implies x-\bar x = 7 - 5 = 2[/tex]

These are sometimes called "residuals".

In the next column, you square these values:

[tex]x=3 \implies (x-\bar x)^2 = (-2)^2 = 4[/tex]

[tex]x=5 \implies (x-\bar x)^2 = 0^2 = 0[/tex]

[tex]x=7 \implies (x-\bar x)^2 = 2^2 = 4[/tex]

and the variance of the data is the sum of these so-called "squared residuals".

In cylindrical coordinates, W is the set of points

W = {(r, θ, z) : 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π and -2 ≤ z ≤ 5}

(A) Then the integral of f(x, y, z) over W is

[tex]\displaystyle\iiint_W(x^2+y^2+z^2)\,\mathrm dV = \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

(B)

[tex]\displaystyle \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^2\int_{-2}^5(r^3+rz^2)\,\mathrm dz\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(zr^3+\frac13rz^3\right)\bigg|_{z=-2}^{z=5}\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(\frac{133}3r+7r^3\right)\,\mathrm dr \\\\\\= 2\pi \left(\frac{133}6r^2+\frac74r^4\right)\bigg|_{r=0}^{r=2} \\\\\\= 2\pi \left(\frac{110}3\right) = \boxed{\frac{220\pi}3}[/tex]

Answer:

I think you have a type.. "the seventh number must be a 1"

there are no 1's in the original set of numbers

Step-by-step explanation:

Answer:

1024

Step-by-step explanation:

Given :-

Value of -2x²y³

Answer:

254

Step-by-step explanation:

^ <- this is the square sign

-2x^y^3

x=2

y=4

put x values in to x place and y value in to y place.

-2(2)^2(4)^3

Find the squares and - it with 2

-2(4)(64)

2-256=254

:. the value of -2x^2y^3 =254

That the answer.

Hope this is what you asked.

The work is simply the dot product of the force and displacement (which I assume are given in Newtons and meters, respectively):

W = F • d

W = (5i + 3j - 2k) N • ((4i - j + 3k) m - (j + k) m)

W = (5i + 3j - 2k) • (4i - 2j + 2k) Nm

W = (20 - 6 - 4) Nm

W = 10 J

Answer:

10

Step-by-step explanation:

Sorry. I needed to answer this question to get access.

[tex]\frac{13}{28}[/tex]

Given:

Blue marbles: 3

Reb marbles: 5

Total marbles: 8

Two marbles are selected at random, one after the other with replacement.

Getting the same colour of marbles from the selection means the two marbles are both red or both blue.

(a) Probability of getting 2 marbles being red in colour

i. Probability of picking a red at the first selection:

Number of red marbles ÷ Total number of marbles

=> 5 ÷ 8 = [tex]\frac{5}{8}[/tex]

ii. Probability of picking a red at the second selection:

Number of remaining red marbles ÷ Total number of remaining marbles

Since after the first pick, the marble is not replaced, the remaining red marbles is 4 while the total number of remaining marbles is 7

=> 4 ÷ 7 = [tex]\frac{4}{7}[/tex]

iii. The probability of getting both marbles being red is the product of i and ii above. i.e

[tex]\frac{5}{8}[/tex] x [tex]\frac{4}{7}[/tex] = [tex]\frac{5}{14}[/tex]

(b) Probability of getting 2 marbles being blue in colour

i. Probability of picking a blue at the first selection:

Number of blue marbles ÷ Total number of marbles

=> 3 ÷ 8 = [tex]\frac{3}{8}[/tex]

ii. Probability of picking a blue at the second selection:

Number of remaining blue marbles ÷ Total number of remaining marbles

Since after the first pick, the marble is not replaced, the remaining blue marbles is 2 while the total number of remaining marbles is 7

=> 2 ÷ 7 = [tex]\frac{2}{7}[/tex]

iii. The probability of getting both marbles being blue is the product of i and ii above. i.e

[tex]\frac{3}{8}[/tex] x [tex]\frac{2}{7}[/tex] = [tex]\frac{3}{28}[/tex]

(c) Probability of getting 2 marbles of the same colour.

The probability of getting 2 marbles of same colour is the sum of the probability of getting both marbles of red colour and the probability of getting both marbles as blue colour. i.e The sum of a(iii) and b(iii)

[tex]\frac{5}{14}[/tex] + [tex]\frac{3}{28}[/tex] = [tex]\frac{13}{28}[/tex]

The probability of getting 2 of the same colour is [tex]\frac{13}{28}[/tex]

Answer:I’m pretty sure ( not 100% thou ) the awnser would be A) 4

Answer:

in a square all sides are equal so x has to equal

128

Hope This Helps!!!

Answer:

79.5, 110.5, 141.5, 172.5, 203.5, 234.5

Step-by-step explanation:

Given

The attached distribution

Required

The lower class limits

To do this, we simply subtract 0.5 from the lower interval

From the attached distribution, the lower intervals are:

80.0, 111.0, 142.0, 173,0 .......

So, the lower class limits are:

[tex]80.0-0.5 = 79.5[/tex]

[tex]111.0-0.5 = 110.5[/tex]

[tex]142.0-0.5 = 141.5[/tex]

[tex]173.0-0.5 = 172.5[/tex]

[tex]204.0-0.5 = 203.5[/tex]

[tex]235.0-0.5 = 234.5[/tex]

Answer:

[tex]\bar x = 107.11[/tex]

[tex]\sigma_x = 31.07[/tex]

Step-by-step explanation:

See comment for complete question

Given

[tex]x: 97\ 178\ 129\ 90\ 75\ 94\ 116\ 100\ 85[/tex]

Solving (a): The sample mean

This is calculated using:

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

So, we have:

[tex]\bar x = \frac{97+ 178+ 129+ 90+ 75+ 94+ 116+ 100+ 85}{9}[/tex]

[tex]\bar x = \frac{964}{9}[/tex]

[tex]\bar x = 107.11[/tex]

Solving (b): The sample standard deviation

This is calculated as:

[tex]\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]

So, we have:

[tex]\sigma_x = \sqrt{\frac{(97 - 107.11)^2 +.............+ (85- 107.11)^2 }{9-1}}[/tex]

[tex]\sigma_x = \sqrt{\frac{7720.8889}{8}}[/tex]

[tex]\sigma_x = \sqrt{965.1111125}[/tex]

[tex]\sigma_x = 31.07[/tex]

Answer:

Step 1

Because the number in front of the bracket is 1 and it is also affected by the negative sign(-),5 is supposed to be negative not positive because (negative by positive is negative)

And since the first step has an error in it,the remaining steps would also be wrong.

Answer:

[tex]X=6.67ft/s[/tex]

Step-by-step explanation:

From the question we are told that:

Height of pole [tex]H_p=15[/tex]

Height  of man [tex]h_m=6ft[/tex]

Speed of Man [tex]\triangle a =4ft/s[/tex]

Distance from pole [tex]d=35ft[/tex]

Let

Distance from pole to man=a

Distance from man to shadow =b

Therefore

 [tex]\frac{a+b}{15}=\frac{b}{6}[/tex]

 [tex]6a+6b=15y[/tex]

 [tex]2a=3b[/tex]

Generally the equation for change in velocity is mathematically given by

 [tex]2(\triangle a)=3(\triangle b )[/tex]

 [tex]2*4=3(\triangle b)[/tex]

 [tex]\triangle a=\frac{8}{3}[/tex]

Since

The speed of the shadow is given as

 [tex]X=\triangle b+\triangle a[/tex]

 [tex]X=4+8/3[/tex]

 [tex]X=6.67ft/s[/tex]

Answer:

The expected number of winning balls that Heather draws is 0.3.

Step-by-step explanation:

The balls are chosen without replacement, 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]

Expected value of the hypergeometric distribution:

The expected value is given by:

[tex]E(X) = \frac{nk}{N}[/tex]

Expected number of blue and green balls:

40 balls, which means that [tex]N = 40[/tex]

2 are chosen, which means that [tex]n = 2[/tex]

25 are blue, which means that [tex]k = 25[/tex]

So

[tex]E(X) = \frac{nk}{N} = \frac{25(2)}{40} = 1.25[/tex]

1.25 balls are expected to be blue and 2 - 1.25 = 0.75 green.

Of the blue balls, 12% are winning.

Of the green balls, 20% are winning.

Calculate the expected number of winning balls that Heather draws.

[tex]E_w = 1.25*0.12 + 0.75*0.2 = 0.3[/tex]

The expected number of winning balls that Heather draws is 0.3.

Answer:

x=−40

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x4−3x8=5

14x+−38x=5

(14x+−38x)=5(Combine Like Terms)

−18x=5

−18x=5

Step 2: Multiply both sides by 8/(-1).

(8−1)*(−18x)=(8−1)*(5)

x=−40

Answer:

x=−40

Hello!

x/4 - 3x/8 = 5

2x - 3x = 40

-x = 40

x = -40

Good luck! :)

im struggling with the same one

9514 1404 393

Answer:

  y -9

Step-by-step explanation:

From 4 years ago until 5 years from now, John will age 9 years. That is, his age 4 years ago is 9 years less than it will be in 5 years.

John's age 4 years ago is y-9 years.

Answer:

[0.25, 2]

Step-by-step explanation:

We have

4t² ≤ 9t-2

subtract 9t-2 from both sides to make this a quadratic

4t²-9t+2 ≤ 0

To solve this, we can solve for 4t²-9t+2=0 and do some guess and check to find which values result in the function being less than 0.

4t²-9t+2=0

We can see that -8 and -1 add up to -9, the coefficient of t, and 4 (the coefficient of t²) and 2 multiply to 8, which is also equal to -8 * -1. Therefore, we can write this as

4t²-8t-t+2=0

4t(t-2)-1(t-2)=0

(4t-1)(t-2)=0

Our zeros are thus t=2 and t = 1/4. Using these zeros, we can set up three zones: t < 1/4, 1/4<t<2, and t>2. We can take one random value from each of these zones and see if it fits the criteria of

4t²-9t+2 ≤ 0

For t<1/4, we can plug in 0. 4(0)²-9(0) + 2 = 2 >0 , so this is not correct

For 1/4<t<2, we can plug 1 in. 4(1)²-9(1) +2 = -3 <0, so this is correct

For t > 2, we can plug 5 in. 4(5)²-9(5) + 2 = 57 > 0, so this is not correct.

Therefore, for 4t^2 ≤ 9t-2 , which can also be written as 4t²-9t+2 ≤ 0, when t is between 1/4 and 2, the inequality is correct. Furthermore, as the sides are equal when t= 1/4 and t=2, this can be written as [0.25, 2]

Answer:

main age = total age/total people

if Main age is = 27

[tex]27 = \frac{ \times }{5} [/tex]

and x = 135

Total age is = 135

then main age is 35

[tex][35 = \frac{y}{6} [/tex]

and y = 210

first main age - second main age = age of the person participating

210 - 135 = 75

HAVE A NİCE DAY

Step-by-step explanation:

GREETINGS FROM TURKEY

Answer:

{-6,0,2,4}

Step-by-step explanation:

Answer:

annualy=$3689.62

semiannually=$3695.27

monthly=$3700.06

weekly=$3700.81

daily=$3701.00

Continuously=$3701.03

Step-by-step explanation:

Given:

P=3000

r=3%

t=7 years

Formula used:

Where,

A represents Accumulated amount

P represents (or) invested amount

r represents interest rate

t represents time in years

n represents accumulated or compounded number of times per year

Solution:

(i)annually

n=1 time per year

[tex]A=3000[1+\frac{0.03}{1} ]^1^(^7^)\\ =3000(1.03)^7\\ =3689.621596\\[/tex]

On approximating the values,

A=$3689.62

(ii)semiannually

n=2 times per year

[tex]A=3000[1+\frac{0.03}{2}^{2(4)} ]\\ =3000[1+0.815]^14\\ =3695.267192[/tex]

On approximating the values,

A=$3695.27

(iii)monthly

n=12 times per year

[tex]A=3000[1+\frac{0.03}{12}^{12(7)} \\ =3000[1+0.0025]^84\\ =3700.0644[/tex]

On approximating,

A=$3700.06

(iv) weekly

n=52 times per year

[tex]A=3000[1+\frac{0.03}{52}]^3^6 \\ =3000(1.23360336)\\ =3700.81003[/tex]

On approximating,

A=$3700.81

(v) daily

n=365 time per year

[tex]A=3000[1+\frac{0.03}{365}]^{365(7)} \\ =3000[1.000082192]^{2555}\\ =3701.002234[/tex]

On approximating the values,

A=$3701.00

(vi) Continuously

[tex]A=Pe^r^t\\ =3000e^{\frac{0.03}{1}(7) }\\ =3000e^{0.21} \\ =3000(1.23367806)\\ =3701.03418\\[/tex]

On approximating the value,

A=$3701.03

Answer:

Attached images

It was just easier for me this way.

Let me know in comments if you have questions.

Step-by-step explanation:

Step-by-step explanation:

it will help u