Lazare is dividing 52.8 ounces of cereal equally into 9 bags. Which is the best way to estimate the amount of cereal in each bag?

Answer:

TOTAL TAX IS: $19.25

TOTAL PRICE IS: $294.25

MONEY LEFT IS: $135.75

Step-by-step explanation:

First, you do a proportion:

?/275 = 7/100

the ?/275 represents how much money in $275

the 7/100 represents 7% (7 hundredths)

then you cross multiply:

275 x 7 = 1925

then you divide:

1925 divided by 100 = 19.25

So $19.25 is the tax amount.

To get the total price, you add:

Tax ($19.25) + Cost($275) = $294.25

$294.25 is the total amount that Anna has to pay.

To get what Anna has left over, you subtract:

Has($430) - Spend($294.25) = $135.75

Anna has $135.75 left.

I hope this helped you a lot!

Bri <3

Answer: -3x-7

Step-by-step explanation:

First, remove the parentheses,

Then. simplify

Answer:

y -5 = (3/4)(x -6)

Step-by-step explanation:

Between the two points, the line has a rise of 6 and a run of 8. Thus, the slope is ...

... m = rise/run = 6/8 = 3/4

The point-slope form of the equation of a line with slope m through point (h, k) is ...

... y - k = m(x - h)

For point B, (h, k) = (6, 5), so filling in the values gives ...

... y - 5 = (3/4)(x - 6)

let's move like the crab, backwards, let's get hmmm EF first

\(\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies o=\sqrt{c^2 - a^2} \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{17}\\ a=\stackrel{adjacent}{10}\\ o=\stackrel{opposite}{EF} \end{cases} \\\\\\ EF=\sqrt{ 17^2 - 10^2}\implies EF=\sqrt{ 189 }\implies EF\approx 13.7 \\\\[-0.35em] ~\dotfill\)

\(\sin( F )=\cfrac{\stackrel{opposite}{10}}{\underset{hypotenuse}{17}} \implies \sin^{-1}(~~\sin( F )~~) =\sin^{-1}\left( \cfrac{10}{17} \right) \\\\\\ F =\sin^{-1}\left( \cfrac{10}{17} \right)\implies F \approx 36^o \\\\[-0.35em] ~\dotfill\\\\ \cos(D )=\cfrac{\stackrel{adjacent}{10}}{\underset{hypotenuse}{17}} \implies \cos^{-1}(~~\cos( D )~~) =\cos^{-1}\left( \cfrac{10}{17} \right) \\\\\\ D =\cos^{-1}\left( \cfrac{10}{17} \right)\implies D \approx 54^o\)

Make sure your calculator is in Degree mode.

Answer:

N+1

Step-by-step explanation:

It begins with = 2

then n+1=3

then n+2=4 and so on it is n+1

By running at least 100 simulations and calculating the variance of the project duration, you can estimate the variance of the project based on the uniformly distributed time intervals for each activity.

To estimate the variance of the project duration using simulation in Excel, you can follow these steps:

1. Create a new Excel worksheet and label columns for activities A, B, and C.

2. In the cells under each activity, use the "RANDBETWEEN" function to generate random numbers within the given intervals. For example, for activity A, the formula would be "=RANDBETWEEN(1,5)".

3. In a separate cell, calculate the total duration of the project by summing the durations of activities A, B, and C. For example, if the durations are in cells A1, B1, and C1, the formula would be "=A1 + B1 + C1".

4. Repeat steps 2 and 3 at least 100 times to generate multiple random durations for the project.

5. Calculate the average duration of the project using the "AVERAGE" function. For example, if the durations are in cells D1:D100, the formula would be "=AVERAGE(D1:D100)".

6. Calculate the variance of the project duration using the "VAR.P" function. For example, if the durations are in cells D1:D100, the formula would be "=VAR.P(D1:D100)".

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Answer:

220

Step-by-step explanation:

If we take 20 common (as in we can take it out and group the rest) , we can say 20 (15 7/9-4 7/9) which is equal to 20 (11). 20x11 = 220.  

The derivative of f(x) = x⋅5x is f'(x) = 10x, which means that the rate of change of the function at any point x is 10 times the value of x at that point.

Using the product rule of differentiation, we can find the derivative of the function f(x) = x⋅5x as follows:

f'(x) = (x)'(5x) + x(5x)'

where (x)' and (5x)' are the derivatives of x and 5x with respect to x, respectively.

(x)' = 1 (the derivative of x with respect to x is 1)

(5x)' = 5 (the derivative of 5x with respect to x is 5)

Substituting these values, we get:

f'(x) = 1⋅5x + x⋅5

Simplifying further, we get:

f'(x) = 5x + 5x

Therefore, f'(x) = 10x.

In conclusion, the derivative of f(x) = x⋅5x is f'(x) = 10x, which means that the rate of change of the function at any point x is 10 times the value of x at that point.

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Answer:

This function would have two distinct real roots.

Step-by-step explanation:

Consider a quadratic function \(f(x) = a\, x^{2} + b\, x + c\).

The determinant of this function would be \(\Delta = b^{2} - 4\, a\, c\).

The (only) two roots of this quadratic function would be:

\(\begin{aligned} x_{1} = \frac{-b - \sqrt{\Delta}}{2\, a}\end{aligned}\), and

\(\begin{aligned} x_{2} = \frac{-b + \sqrt{\Delta}}{2\, a}\end{aligned}\).

Because of the square root \(\sqrt{\Delta}\) in the two expressions, \(x_{1}\) and \(x_{2}\) would take real values if and only if \(\Delta \ge 0\) (determinant is nonnegative.) If \(\Delta < 0\), this quadratic function would not have any real root.

Since the only difference between the two roots \(x_{1}\) and \(x_{2}\) is \((1/a)\, \sqrt{\Delta}\), these two roots would repeat one another if \(\Delta = 0\) (determinant is zero.)

Otherwise, if \(\Delta > 0\), this quadratic function would have two distinct real roots.

Answer:

B.   3,158

Step-by-step explanation:

h(x) = -49x − 125

Let x = -67

h(-67) = -49(-67) − 125

          =3283-125

          = 3158

Answer:

Answer B

Step-by-step explanation:

To find the value of h(-67) for the function h(x) = -49x - 125,

we substitute -67 for x in the function and evaluate it.

h ( - 67 ) = - 49 ( - 67 ) - 125

Now we can simplify the expression:

h ( -67 ) = 3283 - 125

h ( -67 ) = 3158

A graph of the triangle is shown in the image attached below.

The coordinates of its vertices include the following:

A' = (-4, 2).

B' = (-3, 5).

C' = (-6, 5).

In Geometry, a reflection over the y-axis is given by this transformation rule (x, y) → (-x, y). This ultimately implies that, a reflection over the y-axis would maintain the same y-coordinate while the sign of the x-coordinate changes from positive to negative or negative to positive.

By applying a reflection over the y-axis to the coordinates of triangle ABC, we have the following coordinates:

Coordinate A = (-4, 2)   →  Coordinate A' = (-(4), 2) = (-4, 2).

Coordinate B = (3, 5)   →  Coordinate B' = (-(3), 4) = (-3, 5).

Coordinate C = (6, 5)   →  Coordinate C' = (-(6), 2) = (-6, 5).

Next, we would use an online graphing calculator to plot the coordinates of the image as shown in the image attached below.

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The equation that represents n; the number of pages Steph read on Monday is B (52 − 16) × 2 = n

An equation is the statement that illustrates the variables given. In this case, two or more components are taken into consideration to describe the scenario.

The following can be deduced:

Lara read 52 pages.

Jack read 16 fewer pages than Lara.

Steph read 2 times as many pages as Jack.

The pages for Steph will be:

= 2(52 - 16)

In conclusion, the correct option is B.

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Answer:

G. 226 mm

Step-by-step explanation:

C=2(3.14)(36)

Answer:

Second choice, 3

Step-by-step explanation:

a^2 + b^2 = c^2

4^2 + b^2 = 5^2

16 + b^2 = 25

b^2 = 9 (square root both sides)

b = 3

Answer:

x= -8; y= 5

Step-by-step explanation:

5x + 8 ( -3x -19) =0

5x -24x -152 =0

5x -24x= 152

-19x = 152

x= -8

y= -3x -19

y= -3(-8) -19

y= 5

5x + 8y = 0

5(-8) + 8(5) =0

-40 + 40 = 0

It can be easily verified that the CDF F(x) derived above satisfies all these properties, and hence it is a valid CDF.

To find the cumulative distribution function (CDF) of the random variable X, we integrate the probability density function (PDF) g(x) over the range (-∞, x].

For x < 0, P(X ≤ x) = 0 because the range of X is 0 ≤ X ≤ 1.

For 0 ≤ x ≤ 1, we have:

P(X ≤ x) = ∫[0,x] g(t) dt

P(X ≤ x) = ∫[0,x] (2 - t) dt

P(X ≤ x) = [2t - (\(t^2\))/2] evaluated from 0 to x

P(X ≤ x) = 2x - \(x^2\)/2

For x > 1, P(X ≤ x) = 1 because the range of X is 0 ≤ X ≤ 1.

Therefore, the CDF of the random variable X is:

F(x) = 0 for x < 0

F(x) = 2x - \(x^2\)/2 for 0 ≤ x ≤ 1

F(x) = 1 for x > 1

To check that this is a valid CDF, we need to verify that it satisfies the following properties:

F(x) is non-negative for all x.

F(x) is non-decreasing for all x.

F(x) approaches 0 as x approaches negative infinity.

F(x) approaches 1 as x approaches positive infinity.

It can be easily verified that the CDF F(x) derived above satisfies all these properties, and hence it is a valid CDF.

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Full Question ;

If the probability density of a random variable is given by x g(x)= {2-X 0 0 < x <1 1<x<2 elsewhere Compute u and o

Answer:

\( { - 2x}^{2} - {8x}^{3} + 6x \\ {x }^{2} - {8x}^{3} + 6x\)

Step-by-step explanation:

Step-by-step explanation:

hgfhwjnsksjehefebhxjddkdjdjydudsisi

Answer:

5, 8, 11

Step-by-step explanation:

To find the first 3 terms substitute n = 1, 2, 3 into n th term rule, that is

a₁ = 3(1) + 2 = 3 + 2 = 5

a₂ = 3(2) + 2 = 6 + 2 = 8

a₃ = 3(3) + 2 = 9 + 2 = 11

The first 3 terms are 5, 8, 11

Answer:

$12

Step-by-step explanation:

C0-starting price

C1-price after discount

P- precentage

C1=C0-P/100*C0

C1=20-40/100*20

C1=20-8

C1=12

Answer:

500000

Step-by-step explanation:

We need to find the value of the given expression :

\(5^{\dfrac{5}{2}}\times 20^{\dfrac{3}{2}}\div 10^{-2}\)

We can do it as per BODMAS rule.

\(5^{\dfrac{5}{2}}\times (\dfrac{20^{\dfrac{3}{2}}}{10^{-2}})\\\\=5^{\dfrac{5}{2}}\times {20^{\dfrac{3}{2}}}\times {10^{2}}\\\\=5^{\dfrac{5}{2}}\times {20^{\dfrac{3}{2}}}\times 100\\\\=5^{\dfrac{5}{2}}\times {(5\times 4)^{\dfrac{3}{2}}}\times 100\\\\=5^{\dfrac{5}{2}}\times {(5^{\dfrac{3}{2}})\times (4)^{\dfrac{3}{2}}}\times 100\\\\=5^{(\dfrac{5}{2}+\dfrac{3}{2})}\times 2^2^{\dfrac{3}{2}}\times 100\\\\=5^4\times 2^3\times 100\\\\=625\times 8\times 100\\\\=500000\)

So, the required answer is 500000.

The correct answer is 4 km per hour Lashonda would swim if there were no current.

Here, we'll suppose that Lashonda is moving at x km/hr. in the absence of any current.

Speed of the current is 2 km/hr.

Distance swim against current is 5km

Time taken to swim against current is  5 ÷ (x - 2)

Distance swim with current is 15km

Time is taken to swim with current is 15 ÷ (x + 2)

Now,

Given that  if the time taken in both instances are the same, So, the equation will be 5 ÷ (x - 2) = 15 ÷ (x + 2)

After solving for x we will be solving;

⇒ 5 ÷ (x + 2) = 15 ÷ (x - 2)

⇒ 5x + 10 = 15x - 30

⇒ 10 + 30 = 15x - 5x

⇒ 40 = 10x

⇒ x= 40 ÷ 10 = 4 km per hour

Hence, 4 km per hour Lashonda would swim if there were no current.

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3. the answer is 8

6. the answer is 7

Answer:

48/8=6

56/7=8

Step-by-step explanation:

48÷ 8 = 6

56÷ 7 = 8

The distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.

We have,

Acceleration of plane i.e. a = 12 m/s²

And,

Velocity of plane i.e. v = 100 m/s

And,

t₁ = 2s

t₂ = 5s

t₃ = 10s

So,

Now,

Using the equation of motion,

i.e.

S = vt - \(\frac{1}{2}\) at²

Here,

S = Distance,

v = initial velocity,

a = acceleration

t = time taken

Now,

For t₁ = 2s,

Putting values in above equation we get,

S = (100 * 2) - (\(\frac{1}{2}\) * 12 * 2²)

On solving we get,

S = 200 - 24 = 176 m,

So,

Plane will be at 176m distance.

Now,

For t₂ = 5s,

Putting values in above equation we get,

S = (100 * 5) - (\(\frac{1}{2}\) * 12 * 5²)

On solving we get,

S = 500 - 150 = 350 m,

So,

Plane will be at 350m distance.

Now,

For t₃ = 10s,

Putting values in above equation we get,

S = (100 * 10) - (\(\frac{1}{2}\) * 12 * 10²)

On solving we get,

S = 1000 - 600 = 400 m,

So,

Plane will be at 400m distance.

Hence, we can say that the distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.

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Answer:

Laura will need 1/4 cups more chocolate

Step-by-step explanation:

If laura has 2 cups of chocolate to begin with and she needs 2 and 1/4 cups when you subtract you get your answer of 1/4.

Hope this Helps have an awesome day!

Answer:

2 1/2 cups

Step-by-step explanation:

2 1/4 doubled is 4 1/2.

(2*2= 4.      1/4*2=1/2)

recipe calls for 2 1/4 cups

doubled it calls for 4 1/2 cups

she has 2 cups

4 1/2- 2= 2 1/2 cups

she needs 2 1/2 cups

Answer:

3/8 shortening

1/4 salt

1 1/8 granulated sugar

1/4 baking soda (or 3/4 if that is 1 1/2 t from the original recipe)

1/8 eggs

1 c cake flour

1/4 vanilla extract

food coloring

1/8 butter flavoring

When I looked at the chart I only saw dots plotted on 10 and a 43 ish number I didn't see anything in between  23 and 35 so I'm guessing your answer is 0%

Hope this helped and have an amazing day <3

Answer:

I actually believe it is 50% because the middle box is worth 50% of the plot.

Step-by-step explanation:

Hello,

\(f(x) = - 5 {x}^{2} + 3\)

\(h(x) = - 5 {x}^{2} + 3 + 5 = - 5 {x}^{2} + 8\)

Answer:5x-29•4+1/3x

Now put x value

5 (-11/5)-29.4+1/3 (-11/5)

-11-29.4+1/(-33/5)

-202/5+1/(-33/5)

-6691/165

Step-by-step explanation:

The answer I got was -617/15

The convolution product of r(t) and s(t) is y(t) = 2(t+3)u(t+3) - 2(t-3)u(t-3) - 2(t+2)u(t+2) + 2(t-2)u(t-2) - 2(t+1)u(t+1) + 2(t-1)u(t-1) - 2tu(t) + 2(t-1)u(t-1) - 2(t-2)u(t-2) + 2(t-3)u(t-3).

To find the convolution product of r(t) and s(t), we need to evaluate the integral of the product of r(t) and s(t) over the appropriate range. In this case, r(t) = u(t) - u(t-1) and s(t) = 2u(t+3) - 2u(t-3).

To perform the convolution, we substitute the expression for r(t) and s(t) into the integral:

y(t) = ∫[u(τ) - u(τ-1)][2u(t+3-τ) - 2u(t-3-τ)] dτ.

Simplifying this expression, we obtain:

y(t) = 2∫[u(τ) - u(τ-1)][u(t+3-τ) - u(t-3-τ)] dτ.

The next step is to evaluate the integral over the appropriate range. Since the limits of integration depend on the variables involved, we need to consider different cases.

Case 1: t+3 ≥ τ ≥ t-3

In this case, both u(t+3-τ) and u(t-3-τ) are equal to 1, and the integral becomes:

y(t) = 2∫[u(τ) - u(τ-1)] dτ.

Case 2: t+3 ≥ τ > t

In this case, u(t+3-τ) = 1, and u(t-3-τ) = 0, so the integral becomes:

y(t) = 2∫[u(τ) - u(τ-1)] dτ + 2∫u(τ-3) dτ.

Case 3: t > τ ≥ t-3

In this case, u(t+3-τ) = 0, and u(t-3-τ) = 1, so the integral becomes:

y(t) = 2∫[u(τ) - u(τ-1)] dτ - 2∫u(τ-3) dτ.

By evaluating the integrals in each case, we can obtain the expression for y(t) as shown in the main answer.

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