which linear inequality represents the graph below?

A. y < -1/4x-4
B. y < 4x-4
C. y < -1/4x+4
D. y < -4x+4​

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

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:

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.

Answer:

A = 16 ft^2

P = 20 ft

Step-by-step explanation:

P = perimeter

A = area

STEP 1: divide the shape into rectangles

Rectangle 1: 2ft*3ft

Rectangle 2: 2ft*5ft

STEP 2: Find the area of each rectangle

Equation for area of a rectangle = bh

Rectangle 1: b = 2, h = 3

Rectangle 2: b = 2, h = 5

(2 * 3) + (2 * 5)

6 + 10

16 ft^2

Now, we have to find the perimeter

STEP 1:  Find the unknown side lengths.

To find the lengths of the sides not labeled, you have to use the lengths of the sides we already know.

The length of one parallel side is 5, and the length of another parallel side is 2. The length of the unknown side starts at the same place as the top of the side length that is 5, and ends at the top of the side length that is 2. This means that we have to subtract 2 from 5 in order to find the unknown side length.

STEP 2: Add up all the side lengths

P = 2 + 5 + 5 + 2 + 3 + 3

P = 20 ft

Don't forget to label your answers!!

I hope this made sense, it's is a little hard to explain in simple terms without being able to draw, but I hope it helped.

Answer:

{-6,0,2,4}

Step-by-step explanation:

Answer:

11 hours is right answer i hope it will help you

Step-by-step explanation:

it will help u

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:

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.

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:

The probability is P = 0.08

Step-by-step explanation:

We have:

2 pink balls

7 purple balls

6 white balls

So the total number of balls is just:

2 + 7 + 6 = 15

We want to find the probability of randomly picking 3 purple balls (without replacement).

For the first pick:

Here all the balls have the same probability of being drawn from the urn, so the probability of getting a purple one is equal to the quotient between the number of purple balls (7) and the total number of balls (15)

p₁ = 7/15

Second:

Same as before, notice that because the balls are not replaced, now there are 6 purple balls in the urn, and a total of 14 balls, so in this case the probability is:

p₂ = 6/14

third:

Same as before, this time there are 5 purple balls in the urn and 13 balls in total, so here the probability is:

p₃ = 5/13

The joint probability (the probability of these 3 events happening) is equal to the product between the individual probabilities, so we have:

P = p₁*p₂*p₃ = (7/15)*(6/14)*(5/13)  = 0.08

Answer:

hey hi mate

hope you like it

plz mark it as brainliest

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]

[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]

im struggling with the same one

Answer:

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

128

Hope This Helps!!!

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:

[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:

10

Step-by-step explanation:

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

Answer:

Attached images

It was just easier for me this way.

Let me know in comments if you have questions.

Step-by-step explanation:

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:

the answer of r is 8 i hope it will help

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:72 [tex]cm^{2}[/tex]

Solution 1:

Step 1: Find EF use Pythagorean theorem

[tex]EF^{2} = EB^{2} + BF^{2}[/tex]

[tex]EF^{2} = 6^{2} + 6^{2}[/tex]

EF = [tex]\sqrt{6^{2} + 6^{2} }[/tex] = 6[tex]\sqrt{2}[/tex] cm

Step 2: The area of EFGH = [tex]EF^{2}[/tex]= [tex](6\sqrt{2} )^{2}[/tex] = 72

Solution 2: See that the area of EFGH is equal [tex]\frac{1}{2}[/tex] the area of ABCD

The area of ABCD = 12x12 = 144

Thus, the area of EFGH = 144: 2 = 72:)

Have a nice day!

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".

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:

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: