Alex eats 2 pieces of candy. Marisol eats 3 times as many pieces of candy as Alex. Which equation can be used to determine the amount of candy that Marisol eats? Let a represent Alex and m represent Marisol.
m=3a
m=3a+2
m=2a
m=2a+3

Answer:

6 1/2

Step-by-step explanation:

Use the area formula: A=LengthxWidth

Plug in what we know: 24 3/8= 3 3/4x Width

We need to isolate the width; get it by itself.

Do this by (24 3/8)/(3 3/4). The best way is to enter this into a calculator.

Your answer is 6 1/2

Answer:

40m^2 :)

Step-by-step explanation:

Area of a triange = (h·b)/2

Now lets solve :)

4 + 6 = 10

(8·10)/2

80/2

40

Have an amazing day!!

Please rate and mark brainliest!!

Answer:

center: (2,-12)

radius: 9

Step-by-step explanation:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

[tex]h=2\\k=-12\\r=9[/tex]

get r by getting square root of 81

center = (h,k)

Answer:

center: (2,-12)

radius: 9

Step-by-step explanation:

(x-h)^2+(y-k)^2=r^2

h=2

k=-12

r=9

Answer:

the 1.5 goes between -1 and 0 and -3/2 too and -4/3 between the-2 and the 3/2 between the 0 and 1

Step-by-step explanation:

Answer:

Smily face=5 lightning=8

Step-by-step explanation:

5+5=10 8x3=24      24+10=34

8x5=40              5x8=40

Just find the vertex and then compare with graph

#a

Parabola opening upwards

Graph 3

#2

Vertex at (-3,0)

Graph IV

#3

Parabola opening downwards as a is -ve

Graph I

#4

One graph is left

Answer:

a)  graph iii)

b)  graph iv)

c)  graph i)

d)  graph ii)

Step-by-step explanation:

Vertex form of a quadratic equation:  [tex]y = a(x - h)^2 + k[/tex]

where [tex](h, k)[/tex] is the vertex (turning point)

First, determine the vertices of the parabolas by inspection of the graphs:

Next, write each given equation in vertex form and compare to the vertices above.

[tex]\textsf{a)}\quad y=2x^2-8[/tex]

[tex]\textsf{Vertex form}: \quad y=2(x-0)^2-8[/tex]

⇒ Vertex = (0, -8)

Therefore, graph iii)

[tex]\textsf{b)} \quad y=(x+3)^2[/tex]

[tex]\textsf{Vertex form}: \quad y=(x+3)^2+0[/tex]

⇒ Vertex = (-3, 0)

Therefore, graph iv)

[tex]\textsf{c)} \quad y=-2|x-4|^2+8[/tex]

[tex]\textsf{Vertex form}: \quad y=-2|x-4|^2+8[/tex]

⇒ Vertex = (4, 8)

Therefore, graph i)

[tex]\textsf{d)} \quad y=(x-3)^2-8[/tex]

[tex]\textsf{Vertex form}: \quad y=(x-3)^2-8[/tex]

⇒ Vertex = (3, -8)

Therefore, graph ii)

Answer:

using [tex]\pi[/tex]:   65.45 in³ (nearest hundredth)

using [tex]\pi =3.14[/tex]:   65.42 in³ (nearest hundredth)

Step-by-step explanation:

The radius of the sphere is half the side length of the cube (see attached diagram).  Therefore, the side length of the cube = 2r

Given:

[tex]\textsf{Volume of a cube}=x^3\quad \textsf{(where}\:x\:\textsf{is the side length)}[/tex]

[tex]\implies 125=(2r)^3[/tex]

[tex]\implies \sqrt[3]{125}=2r[/tex]

[tex]\implies 5=2r[/tex]

[tex]\implies r=\dfrac52[/tex]

Substitute the found value of r into the volume of a sphere equation:

[tex]\begin{aligned}\textsf{Volume of a sphere} & =\dfrac43 \pi r^3\\\\ & =\dfrac43 \pi \left(\dfrac52\right)^3\\\\ & =\dfrac43 \pi \left(\dfrac{125}{8}\right)\\\\ & =\dfrac{500}{24} \pi\\\\ & =\dfrac{125}{6} \pi\\\\ & =65.45\:\sf in^3\:(nearest\:hundredth) \end{aligned}[/tex]

Inequalities are expressions not separated by an equal sign. The solution of the compound inequality will be -6 < x < 24

Inequalities are expressions not separated by an equal sign.

Given the compound inequalities

X - 7 < 17 or -6x >36

For the inequality

x - 7 < 17

x < 17 + 7

x < 24

Similarly for -6x >36

x < -36/6
x < -6

The solution of the compound inequality will be -6 < x < 24

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When the variances are known and the sample size is high, a z-test is used to assess if two population means vary. The mean of the two players is not the same.

When the variances are known and the sample size is high, a z-test is used to assess if two population means vary.

In order to execute an appropriate z-test, the test statistic is expected to have a normal distribution, and nuisance factors such as standard deviation should be known.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

[tex]\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ H_a: \mu_1 & \ne & \mu_2 \end{array}[/tex]

This corresponds to a two-tailed test, and a z-test for two means, with known population standard deviations, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is [tex]z_c = 1.96[/tex]

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

[tex]\begin{array}{ccl} z & = & \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ {\sigma_1^2/n_1} + {\sigma_2^2/n_2} }} \\\\ & = & \displaystyle \frac{ 30.1 - 25.5}{\sqrt{ {9.112^2/70} + {12.602^2/70} }} \\\\ & = & 2.475 \end{array}[/tex]

[tex]\begin{array}{ccl} z & = & \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ {\sigma_1^2/n_1} + {\sigma_2^2/n_2} }} \\\\ & = & \displaystyle \frac{ 30.1 - 25.5}{\sqrt{ {9.112^2/70} + {12.602^2/70} }} \\\\ & = & 2.475 \end{array}[/tex]

(4.) The decision about the null hypothesis

Since it is observed that |z| = 2.475 > z_c = 1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0133p=0.0133, and since p = 0.0133 < 0.05p=0.0133<0.05, it is concluded that the null hypothesis is rejected.

(5.) As it is concluded that the null hypothesis H₀ is rejected. Therefore, there is enough evidence to claim that the population mean μ1  is different than μ2.

Hence, the mean of the two players is not the same.

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Answer:

If each tile has a number from 0 thru 9 then the possible outcomes for each tile are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Answer:

Step-by-step explanation:

Length of Fence

Cost

Step-by-step explanation:

Tangent is equal to sine over cosine. Cotangent is equal to cosine over sine. Therefore:

[tex]cos(x)( \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} )[/tex]

Distribute the cos(x) into the sum to get:

[tex]sin(x) + \frac{ {cos(x)}^{2} }{sin(x)} [/tex]

Get a common denominator by multiplying the first term by sine over sine to get:

[tex] \frac{ {sin(x)}^{2} }{sin(x)} + \frac{ {cos(x)}^{2} }{sin(x)} [/tex]

The numerator adds to equal 1 due to a common trigonometric identity. Therefore the only remaining term is:

[tex] \frac{1}{sin(x)} [/tex]

Answer:

I have graphed them on desmos and given you three options to choose from:

Step-by-step explanation:

1. with lines (complete shape)

2. with lines (incomplete shape)

3. without lines

Hope this helps!

Answer:

here

Step-by-step explanation:

Answer:

7290in ^2

Step-by-step explanation:

covert 30yd to feet which equals too 90ft.

Then multiple 90 x 81 = 7290

Answer:

30

Step-by-step explanation:

10 x 3 = 30

Answer:

30 centimeters squared

Step-by-step explanation:

The area of a parallelogram can be found by using the formula base x height. The base of this parallelogram is 10, and the height is 3. Therefore the area is 30 square centimeters.

Jennifer idea of  roping off more total area if the tie two ropes together to make one giant BBQ  is a true statement.

The total surface area is known to be the addition of all the sum of any given area of surfaces of a solid together.

Conclusively, Note that by joining together the 2 parties, one can cut off some activities that will be be doubled if it were two separate event and as such, the use of space is been maximize and then the area of the space they will use for the event will be reduced.

Learn more about  total surface area from

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Answer:

[tex]\displaystyle \frac{1}{4}[/tex]

Step-by-step explanation:

Given:

[tex]\displaystyle\frac{4^{6} }{4^{7}}[/tex]

Expand:

[tex]\displaystyle\frac{4*4*4*4*4*4 }{4*4*4*4*4*4*4}[/tex]

Simplify ones, in this case [tex]\frac{4}{4}[/tex]:

[tex]\displaystyle \frac{1}{4}[/tex]

     If you raise 4 to the power of 6, to the power of 7, and then simplify the fraction, you will get the same result. This method is much quicker when you have numbers with the same bases (here the base is 4) so I used this way.

Answer:

576cm³

Step-by-step explanation:

8×6=48

48×12=576

Answer:

-1/12

Step-by-step explanation:

1/4 - 1/3 =   3/12 - 4/12   = - 1/12

Answer:

62cm

Step-by-step explanation:

You were told the perimeter of the triangle is 38

but two sides are added to make a Pentagon. and that Pentagon is made up of a triangle and a square.

so if we find the side of the square, we can equally get the other two sides of the triangle.

hint: (because it's a square, we would be getting an isosceles triangle, meaning two sides are equal)

because one side of the pentagon added to the other two sides of the triangles gives us the perimeter of the triangle.

we first find the sides of the square with all sides equal by diving by 2

[tex] \frac{48}{2} = 12[/tex]

the sides of the square making the perimeter of the square 12cm.

now we subtract 12 from the perimeter of the triangle 38

[tex]38 - 12 = 26[/tex]

now that we have gotten one side we can get the other two by dividing the result by 2

[tex] \frac{26}{2} = 13[/tex]

so the other two sides of the triangle are 13cm.

now since we have the sides of the square and the triangle, we can find the perimeter of the pentagon by adding.

all the sides of the triangle without the base which is also the top of the square.

so instead of adding 4 times of 12, we add 3 times of 12 and 2 times of 13.

3(12)+2(13) = 62

or

12 + 12 + 12 + 13 + 13 = 62cm

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

In this problem:

The probability that at least 5 received a busy signal is given by:

[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009[/tex]

[tex]P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062[/tex]

[tex]P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218[/tex]

[tex]P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513[/tex]

[tex]P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903[/tex]

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295[/tex]

0.8295 = 82.95% probability that at least 5 received a busy signal.

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Answer:

Step-by-step explanation:

4a - (4a-3)

= 4a -1(4a - 3)

= 4a - 4a + 3    <--- Distributing the -1 over the parenthes

= 3       <---- ( as 4a - 4a = 0)

Answer:

Its B

Step-by-step explanation:

Answer:

1

I've attached a screenshot from my graphing calculator of [tex]f(x)[/tex] in blue and [tex]f^{-1}(x)[/tex] in red. Notice how the red line (inverse function) has (3, 1)

Step-by-step explanation:

[tex]f(x) = 3^x\\f^{-1}(x) =\ ?\\[/tex]

We must first figure out the inverse function of [tex]f(x)[/tex] which is  [tex]f^{-1}(x)[/tex]

[tex]y = f(x) = 3^x\\y = 3^x\\log\ y = log\ 3^x\\log\ y = x \times log\ 3\\x = \frac{log\ y}{log\ 3}\\[/tex]

We say y = f(x) to begin with, but after find x = ...

we must 'swap' x and y

[tex]x = \frac{log\ y}{log\ 3}\\y = \frac{log\ x}{log\ 3}\\[/tex]

[tex]f^{-1}(x) = \frac{log\ x}{log\ 3}[/tex]

[tex]f^{-1}(3) = \frac{log\ 3}{log\ 3} = 1[/tex]

FGHIJ ≅ STUQR

Solve for HI

[tex]\sf \dfrac{HI}{GH} = \dfrac{UQ}{UT}[/tex]

[tex]\rightarrow \sf \dfrac{HI}{56} = \dfrac{25}{35}[/tex]

[tex]\rightarrow \sf HI = 40[/tex]

Solve for QR

[tex]\sf \dfrac{IJ}{GH} = \dfrac{QR}{UT}[/tex]

[tex]\rightarrow \sf \dfrac{72}{56} = \dfrac{QR}{35}[/tex]

[tex]\rightarrow \sf QR = 45[/tex]

Solve for ST

[tex]\sf \dfrac{FG}{GH} = \dfrac{ST}{UT}[/tex]

[tex]\rightarrow \sf \dfrac{48}{56} = \dfrac{ST}{35}[/tex]

[tex]\rightarrow \sf ST =30[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{GH}{HI}=\dfrac{UT}{UQ}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{56}{HI}=\dfrac{35}{25}=\dfrac{7}{5}[/tex]

[tex]\\ \rm\Rrightarrow HI=56(5)/7=8(5)=40[/tex]

#QR

[tex]\\ \rm\Rrightarrow \dfrac{GH}{JI}=\dfrac{UT}{RQ}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{56}{72}=\dfrac{35}{RQ}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{7}{9}=\dfrac{35}{RQ}[/tex]

[tex]\\ \rm\Rrightarrow RQ=35(9)/7=5(9)=45[/tex]

#ST

[tex]\\ \rm\Rrightarrow \dfrac{GH}{FG}=\dfrac{UT}{ST}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{56}{48}=\dfrac{7}{6}=\dfrac{35}{ST}[/tex]

[tex]\\ \rm\Rrightarrow ST=6(35)/7=6(5)=30[/tex]

[tex]\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill &75000\\ P=\textit{initial amount}\dotfill &400000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &39\\ \end{cases} \\\\\\ 75000=400000(1 - \frac{r}{100})^{39}\implies \cfrac{75000}{400000}=\left( \cfrac{100-r}{100} \right)^{39}[/tex]

[tex]\cfrac{3}{16}=\left( \cfrac{100-r}{100} \right)^{39}\implies \sqrt[39]{\cfrac{3}{16}}=\cfrac{100-r}{100}\implies 100\sqrt[39]{\cfrac{3}{16}}=100-r \\\\\\ r=100-100\sqrt[39]{\cfrac{3}{16}}\implies r\approx 4.2[/tex]