At that rate, about how many gallons of milk will they need to purchase in a year’s time?

Give your answer as a whole number.

Answer:

Both the numbers are 8.

Step-by-step explanation:

Let the two numbers are p and 64/p.

The sum is given by

[tex]S = p +\frac{64}{p}\\\\\frac{dS}{dp}= 1 - \frac{64}{p^2}\\\\\frac{dS}{dp}=0\\\\\frac{64}{p^2}=1\\\\p= \pm 8[/tex]

So, the sum is minimum for p = 8 0r - 8, so the two numbers 8.

Answer:

11/12 cups

Step-by-step explanation:

2/3+1/4 = ( 2x4 + 3x1 )/( 3x4 ) = ( 8+3 )/12 = 11/12

Answer:

NO

Step-by-step explanation:

NO

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:

yes

Step-by-step explanation:

but I can't see them here

Answer:

75%

Step-by-step explanation:

Given that the score of 8 students in Mr. M's class are 87, 55, 96, 38, 83, 64, 44, 81, the scores less than 84 are 55, 38, 83, 64, 44, 81.

These means that 6 student had scores less that 84 of the 8 students hence the percentage of these test scores that are less than 84

= 6/8 * 100%

= 75%

This means that 75% of the students had scores less than 84

Answer:

Probability[Number of people from church] = 0.26 (Approx.)

Step-by-step explanation:

Given:

Total number of adult in survey = 1,033

Missing information:

Number of people from church = 269

Find:

Probability[Number of people from church]

Computation:

Probability of an event = Number of favourable outcomes / Number of total outcomes

Probability[Number of people from church] = Number of people from church / Total number of adult in survey

Probability[Number of people from church] = 269 / 1,033

Probability[Number of people from church] = 0.2604

Probability[Number of people from church] = 0.26 (Approx.)

9514 1404 393

Answer:

  b. No; the domain values are not at regular intervals although the range values have a common factor.

Step-by-step explanation:

The differences between x-values are ...

  -1, -1, -1, -2 . . . . not a constant difference

The ratios of y-values are ...

  2/8 = 0.5/2 = 0.125/0.5 = 0.25 . . . . a constant difference

The fact that the domain values do not have a common difference renders the common factor of the range values irrelevant. The relation is not exponential.

Answer:

0.4204 probability, option b.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

[tex]f(x) = \mu e^{-\mu x}[/tex]

In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.

The probability that x is lower or equal to a is given by:

[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]

Which has the following solution:

[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]

The probability of finding a value higher than x is:

[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]

Children arrive at a house to do Halloween trick-or- treating according to a Poisson process at the unlucky rate of 13/hour

13 arrivals during an hour, which means that the mean time between arrivals, in minutes is of [tex]\mu = \frac{13}{60} = 0.2167[/tex]

What is the probability that the time between the 15th and 16th arrivals will be more than 4 minutes ?

This is P(X > 4). So

[tex]P(X > 4) = e^{-0.2167*4} = 0.4204[/tex]

So the correct answer is given by option b.

Answer:

it is 61-28 but I not sure u can scan for any application to make sure u get it ur answer thx for

Answer:

10

Step-by-step explanation:

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

Answer:

[tex]( \frac{1}{3})^{3} [/tex]

Step-by-step explanation:

There are many expressions that can be equivalent to 1/27.

For example, 2/54, 3/81 etc

But I think the expression you are looking for is

[tex] \frac{1}{27} = \frac{1 \times 1 \times 1}{3 \times 3 \times 3} = \frac{ {1}^{3} }{ {3}^{3} } = ( \frac{1}{3} )^{3} [/tex]

Hope this is helpful

Answer:

28

Step-by-step explanation:

We know that ,

Answer:

80b-11

Step-by-step explanation:

14b + 1 - 6(2 - 11b)

Distribute

14b+1-12+66b

Combine like terms

80b-11

Answer:

80b - 11

Step-by-step explanation:

what is the problem ?

just multiply it out and combine terms.

14b + 1 - 6(2 - 11b) = 14b + 1 - 12 + 66b = 80b - 11

Answer:

Step-by-step explanation:

Answer:

The answer is A

Step-by-step explanation:

Hope this helps

Answer:

$35,000

Step-by-step explanation:

if $50,000 is to install an area of 1,000 square feet swimming pool and $35,000 can be used to install an 800 square foot swimming pool I think the best graph model is 800 square feet for $35,000 for a cost cut of $15,000 is a good bargain

Answer:

1) subtracting 5

2) adding 20

3) dividing by 2 (multiplying by 1/2)

4) multiplying by 1/10 (dividing by 10)

Step-by-step explanation:

There are four main operations in math: adding, subtracting, multiplying, and dividing. Each of the operations has an opposite. Adding and subtracting are opposites and multiplying and dividing are opposites. This means that subtracting can undo adding and vice versa; additionally, dividing can undo multiplying or vice versa. So, to find the opposite of something switch the operation to the opposite and keep the number. However, it is important to note that with multiplying and dividing you can also find the opposite by keeping the operation while changing the number to the reciprocal.

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:

c....................

Answer:

79.8

Step-by-step explanation:

math

9514 1404 393

Answer:

  y = 2

Step-by-step explanation:

The figure must be flipped top-to-bottom, so the line of reflection must be a horizontal line. Point B must be reflected to itself, so it is on the line of reflection. That means the line of reflection is y = 2.

__

You can draw the line of reflection using any two points that have y-coordinates of 2, for example, (0, 2) and (2, 2).

Answer:

7 digits can be used for each position

There are a total of 5 positions

N = 7^5 = 16,807 numbers

You have 7 choices for the first position, second position, etc.

Answer:

[tex]3.8\:\mathrm{m/s}[/tex]

Step-by-step explanation:

Use the formula [tex]d=rt[/tex] (distance is equal to rate/speed multiplied by time) to solve this problem.

We know that one minute is equal to 60 seconds. Therefore, the distance travelled by the cat in 1 minute is equal to [tex]d=3\cdot 60=180\text{ meters}[/tex].

To catch the cat, the dog needs to also cover an additional 48 meters, because the cat was initially 48 meters away from the dog and it ran away from the dog. Hence, the dog will need to cover [tex]180+48=228[/tex] meters in one minute.

Therefore, we have:

[tex]228=60r,\\r=\frac{228}{60}=\boxed{3.8\:\mathrm{m/s}}[/tex]

Answer:

[tex] \boxed{3.8 \: m/s} [/tex]

Explanation

The first step is to set the speed and the distance equal to the unknown rate of the dog.

3 m/s + 48 m = x m/60s.

Then substitute 60s in for both rates to get distance.

180m + 48m = x m/60

228m = 60x m

÷60 ÷60

3.8m = x m/s.

x = 3.8m/s

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:

The system has an infinite set of solutions [tex](x,y) = (x, \frac{x-5}{4})[/tex]

Step-by-step explanation:

From the first equation:

[tex]20x - 80y = 100[/tex]

[tex]20x = 100 + 80y[/tex]

[tex]x = \frac{100 + 80y}{20}[/tex]

[tex]x = 5 + 4y[/tex]

Replacing on the second equation:

[tex]-14x + 56y = -70[/tex]

[tex]-14(5 + 4y) + 56y = -70[/tex]

[tex]-70 - 56y + 56y = -70[/tex]

[tex]0 = 0[/tex]

This means that the system has an infinite number of solutions, considering:

[tex]x = 5 + 4y[/tex]

[tex]4y = x - 5[/tex]

[tex]y = \frac{x - 5}{4}[/tex]

The system has an infinite set of solutions [tex](x,y) = (x, \frac{x-5}{4})[/tex]

Answer:

the number of flight lines needed is approximately 72

Step-by-step explanation:

Given the data in the question;

Aerial photography is to be taken of a tract of land that is 8 x 8 mi²

L × B = 8 x 8 mi²

Flying height H = 4000 ft = ( 4000 × 12 )inches = 48000 in

focal length f = 6 in

[tex]l[/tex] × b = 9 × 9 in²

side overlap = 30% = 0.3

meaning remaining side overlap = 100% - 30% = 70% = 0.7

{ not end to end overlap }

we take 100% { remaining overlap }

[tex]l[/tex]' = 9 × 100% = 9 in

b' = 9 × 70% = 6.3 in

Now the scale will be;

Scale = f/H

we substitute

Scale = 6 in /  48000 in = 1 / 8000

so our scale is; 1 : 8000

⇒ 1 in = 8000 in

⇒ 1 in = (8000 / 63360)mi

⇒ 1 in = 0.126 mi

so since

L × B = 8 x 8 mi²

[tex]l[/tex]' = ( 9 × 0.126 mi ) = 1.134 mi

b' = ( 6.3 × 0.126 mi ) = 0.7938 mi

Now we get the flight lines;

N = ( L × B ) / ( [tex]l[/tex]' × b' )

we substitute

N = ( 8 mi × 8 mi ) / ( 1.134 mi × 0.7938 mi )

N = 64 / 0.9001692

N = 71.0977 ≈ 72

Therefore, the number of flight lines needed is approximately 72

Answer:

The value of 24x0.03 is gong to be, "800"