I can’t figure it out!! Please help!

Answer:

[tex]Total = 4.8 * 10^{10}[/tex]

Step-by-step explanation:

Given

[tex]h = 1.2 * 10^8[/tex] --- households

[tex]g = 400[/tex] --- gallons

Required

The number of households

To do this, we simply multiply the average households by the gallons.

[tex]Total = g* h[/tex]

[tex]Total = 400 * 1.2 * 10^8[/tex]

[tex]Total = 480 * 10^8[/tex]

Rewrite as:

[tex]Total = 4.8 * 10^2 * 10^8[/tex]

[tex]Total = 4.8 * 10^{10}[/tex]

Answer:

0.1848 = 18.48% probability that more than 4 customers arrive in a 10 minute interval.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Rate of 18 customers per hour.

This is [tex]\mu = 18n[/tex], in which n is the number of hours.

10 minute interval:

An hour has 60 minutes, so this means that [tex]n = \frac{10}{60} = \frac{1}{6}[/tex], and thus [tex]\mu = 18\frac{1}{6} = 3[/tex]

What is the probability that more than 4 customers arrive in a 10 minute interval?

This is:

[tex]P(X > 4) = 1 - P(X \leq 4)[/tex]

In which:

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

Then

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]

[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]

[tex]P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240[/tex]

[tex]P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240[/tex]

[tex]P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680[/tex]

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex] = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 = 0.8152[/tex]

And

[tex]P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8152 = 0.1848[/tex]

0.1848 = 18.48% probability that more than 4 customers arrive in a 10 minute interval.

Answer:

b + 79 =180

b =180-79

=101

Answer:

6144π ft³ ; 19292.2 ft³

Step-by-step explanation:

The volume of the cylinder Given above :

Volume of cylinder, V = πr²h

r =Radius = 16 ; h = 24 ft

V = π * 16² * 24

V = 256 * 24 * π

V = 6144π

Using π = 3.14

V = 6144 * 3.14 = 19292.16

Answer:

y = 1/2x +5

Step-by-step explanation:

change f(x) to y

y = 2x - 10

then switch x and y

x = 2y - 10

then solve for y

add 10 to both sides and divide both sides by 2

Answer:

Step-by-step explanation:

One way to find the inverse is:

1.  replace the symbol f(x) with y (for simplicy--stay tuned).  Now you have

  y = 2x -10

2.  switch  x  and  y.  You get  x = 2y - 10

3.  Solve for y.

x + 10 = 2y

y = (x + 10)/2

4.  Replace  y  with the symbol for the inverse function,

Another approach is to think about inverse operations in reverse order.

f(x): start with x

multiply by 2

subtract 10

Inverse: start with x

add 10 (addition is the inverse of subtraction)

divide by 2 (division is the inverse operation of multiplication)

Answer:

ans : option 2nd

Step-by-step explanation:

total angle substand perimeter of circle so,

so solve by using unitary methods

Answer:

$32.50

Step-by-step explanation:

The total amount of candy bars is 26. Half of that is 13. 13 times the $2.50 for each is $32.50.

Answer:

A ≈ 26.6° (nearest tenth)

Step-by-step explanation:

The diagram shows a right triangle. To solve for A, we would apply trigonometric ratio formula.

Reference angle = <A

Opposite = 6 m

Adjacent = 12 m

Apply TOA

Tan A = Opp/Adj

Substitute

Tan A = 6/12

Tan A = 0.5

A = [tex] Tan^{-1}(0.5) [/tex]

A ≈ 26.6° (nearest tenth)

Answer:  roughly 151.81788 square meters of metal

=====================================================

Explanation:

The base is a square with side lengths x, so its area is x*x = x^2

Let h be the height of the tank. We have four identical wall panels that have area of xh square meters. The four walls lead to a lateral surface area of 4xh. Overall, the entire tank requires x^2+4xh square meters of metal. We're ignoring the top since the tank is open.

-----------

Let's set up a volume equation and then isolate h.

volume = length*width*height

108 = x*x*h

108 = x^2*h

x^2*h = 108

h = 108/(x^2)

-----------

Plug that into the expression we found at the end of the first section.

x^2+4xh

x^2+4x(180/(x^2))

x^2+(720/x)

------------

Depending on what class you're in, the next step here will vary. If you are in calculus, then use the derivative to determine that the local min happens at approximately (7.11379, 151.81788)

If you're not in calculus, then use your graphing calculator's "min" feature to locate the lowest point on the f(x) = x^2+(720/x) curve.

This lowest point tells us what x must be to make x^2+(720/x) to be as small as possible, where x > 0.

In this context, it means that if the square base has sides approximately 7.11379 meters, then you'll need roughly 151.81788 square meters of metal to form the open tank. This is the least amount of metal required to build such a tank, and that will have a volume of 108 cubic meters.

Answer:

Fail to reject the null hypothesis

[tex]CI = (-4.278, 0.478)[/tex]

Step-by-step explanation:

Given

[tex]n_1=n_2 = 18[/tex]

[tex]\bar x_1 = 12.7[/tex]    [tex]\bar x_2 = 14.6[/tex]

[tex]\sigma_1 = 3.2[/tex]    [tex]\sigma_2 = 3.8[/tex]

[tex]\alpha = 0.05[/tex]

Solving (a): Test the hypothesis

We have:

[tex]H_o : \mu_1 - \mu_2 = 0[/tex]

[tex]H_a : u1 - u2 \ne 0[/tex]

Calculate the pooled standard deviation

[tex]s_p = \sqrt\frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2}{n_1+n_2-2}}[/tex]

[tex]s_p = \sqrt\frac{(18-1)*3.2^2 + (18-1)*3.8^2}{18+18-2}}[/tex]

[tex]s_p = \sqrt\frac{419.56}{34}}[/tex]

[tex]s_p = \sqrt{12.34}[/tex]

[tex]s_p = 3.51[/tex]

Calculate test statistic

[tex]t = \frac{x_1 - x_2}{s_p*\sqrt{1/n_1 + 1/n_2}}[/tex]

[tex]t = \frac{12.7 - 14.6}{3.51 *\sqrt{1/18 + 1/18}}[/tex]

[tex]t = \frac{-1.9}{3.51 *\sqrt{1/9}}[/tex]

[tex]t = \frac{-1.9}{3.51 *1/3}[/tex]

[tex]t = \frac{-1.9}{1.17}[/tex]

[tex]t = -1.62[/tex]

From the t table, the p value is:

[tex]p = 0.114472[/tex]

[tex]p > \alpha[/tex]

i.e.

[tex]0.114472 > 0.05[/tex]

So, the conclusion is that: we fail to reject the null hypothesis.

Solving (b): Construct 95% degree freedom

[tex]\alpha = 0.05[/tex]

Calculate the degree of freedom

[tex]df = n_1 + n_2 - 2[/tex]

[tex]df = 18+18 - 2[/tex]

[tex]df = 34[/tex]

From the student t table, the t value is:

[tex]t = 2.032244[/tex]

The confidence interval is calculated as:

[tex]CI = (x_1 - x_2) \± s_p * t * \sqrt{1/n_1 + 1/n_2}[/tex]

[tex]CI = (12.7 - 14.6) \± 3.51 * 2.032244 * \sqrt{1/18 + 1/18}[/tex]

[tex]CI = (12.7 - 14.6) \± 3.51 * 2.032244 * \sqrt{1/9}[/tex]

[tex]CI = (12.7 - 14.6) \± 3.51 * 2.032244 * 1/3[/tex]

[tex]CI = -1.90 \± 2.378[/tex]

Split

[tex]CI = (-1.90 - 2.378, -1.90 + 2.378)[/tex]

[tex]CI = (-4.278, 0.478)[/tex]

Answer and Step-by-step explanation:

[tex]\frac{3}{t}[/tex] + 5

Above is the answer.

#teamtrees #PAW (Plant And Water)

Answer:

H0 : μd = 0

H1 : μd > 0

Step-by-step explanation:

The scenario described above can be compared statistically using a paired test mean as the mean if the two groups are dependent, the two restaurants, Albuquerque and Santa Fe are both restaurant locations of a single restaurant company. Hence, to test the mean difference, we use the paired test statistic. Defined thus `

Null hypothesis ; H0 : μd = 0 and the Alternative hypothesis ; H1 : μd > 0

Answer:

Two Sample T test with Ha = Albernuque>Santa Fe

Step-by-step explanation:

Khan

Answer:

( 1.  6 )

Step-by-step explanation:

Use the formula:

x = - b/2a

to find the maximum and minimum.

Answer: (1,  6 )

Step-by-step explanation:

This should help:

x = - b/2a

Answer:

f(x)= -25x+4

y-inter x=0

y= -25(0)+4

=4

x-inter y=0

0= -25x+4

-4= -25x

x=4/25

[tex]{ \bf{ \underbrace{Given :}}}[/tex]

Radius of the circle "[tex]r[/tex]" = 9 in.

[tex]{ \bf{ \underbrace{To\:find:}}}[/tex]

The circumference of the circle.

[tex]{ \bf{ \underbrace{Solution :}}}[/tex]

[tex]\sf\orange{The\:circumference \:of\:the\:circle\:is\:56.52\:in.}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}[/tex]

We know that,

[tex]\sf\purple{Circumference\:of\:a\:circle \:=\:2πr }[/tex]

[tex] = 2 \times 3.14 \times 9 \: in \\ \\ = 56.52 \: in[/tex]

Therefore, the circumference of the circle is 56.52 in.

[tex]\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{♡}}}}}[/tex]

[tex] \huge \underline \mathcal{Answer}[/tex]

The given angles forms linear pair, and we know the angles forming linear pair are supplementary,

Therefore,

Angle MHJ + Angle MHL = 180°

Let's solve :

Value of variable m = 10°

[tex] \mathrm{✌TeeNForeveR✌}[/tex]

Answer:

Step-by-step explanation:

5 + x - 12 = x- 7 (Add the 5 and -12 to simplify)

x - 7 = x - 7        (notice its the same on both sides of equal sign. Add 7 to both                    sides)

x = x

solution is all real numbers

Part B

5 - 4 - 12 = -4 - 7

-11 = -11

5 + 0 - 12 = 0 - 7

-7 = -7

5 + 5 - 12 = 5 - 7

-2 = -2

Answer:

Step-by-step explanation:

(B). 2√2

Answer:

A 180-pound kangaroo's recommended height is 6 feet.

Step-by-step explanation:

Since 120-pound kangaroo is 5 feet tall and for every inch over this height we add 5 pounds to the total weight, we can make this formula. [tex]120+x=180[/tex]. Using this we can figure out the weight difference between a 120-pound kangaroo and a 180-pound kangaroo. After solving for x we get [tex]x=60[/tex].

Now that we know the 180-pound kangaroo weighs 60 more pounds than a 120-pound kangaroo we can divide that weight difference by 5 to figure out the height difference. [tex]60/5=12[/tex]

After getting the height difference from the weight difference we can conclude that a 180-pound kangaroo's recommended height is 5 feet and 12 inches. Since 12 inches = 1 foot we can easily convert 5' 12" to 6'. This means that a 180-pound kangaroo is 6 feet tall.

Answer:

We know that the height equation is given by:

H(t) = -16*t^2 + 108*t + 28

in ft.

First, we want to find the maximum height of the ball.

The first thing we can see is that the leading coefficient of the quadratic equation is negative, this means that the arms of the graph will open downwards, so the vertex of the quadratic equation is the maximum.

We also know that the ball will reach its maximum height when its velocity is zero (this means that the object stops going upwards at this point).

To get the velocity equation we need to derivate the above equation, we will get:

V(t) = 2*(-16)*t + 1*108

V(t) = -32*t + 108

We need to find the value of t such that this is zero, we will get:

V(t) = 0 =  -32*t + 108

        32*t = 108

             t = 108/32 = 3.375

So the ball reaches its maximum height after 3.375 seconds.

Then the maximum height is given by the height equation evaluated in that time, we will get:

H(3.375) =  -16*(3.375)^2 + 108*3.375 + 28 = 210.25

Then the maximum height of the ball is 210.25 ft

The ball will hit the ground when:

H(t) = 0

Then we just need to solve:

0 =  -16*t^2 + 108*t + 28

Using the Bhaskara's equation we can find that the two solutions for t are:

[tex]t = \frac{-108 \pm \sqrt{(108)^2 - 4*(-16)*28} }{2*(-16)} = \frac{-108 \pm 116}{-32}[/tex]

So the two solutions are:

t = (-108 + 116)/-32 = -0.25

t = (-108 - 116)/-32 = 7

Because t represents time, we should take only the positive value of time (as t = 0 is the time when the ball is thrown).

Then we can conclude that the ball hits the ground after 7 seconds.

Answer:

you first use the coefficients of the letters to times each equation.

Then you either subtract or divide to eliminate the first letter and it coeeffient.

then whatever you get you will divide it with the one with no letter.

then you substitute the answer for the letter in any of the equation.

then you solve it.

I hope this is helpful.

Answer:

40%

Step-by-step explanation:

Because you want to know the total percent the price increased so you add the amounts. 30% plus 10% makes 40%

The sum of the measures of angle LMN and angle NMP is 180 degrees. The measure of ∠LMN is 153°.

In Euclidean geometry, an angle is a shape created by two rays that share a terminus and are referred to as the angle's sides and vertices, respectively.

L, m n is a straight angle, and we want to find the measure of angle, l, m, p, and m p, and since we are told that l m n is a straight angle.

∠LMN + ∠NMP = 180°

The straight line is 180°

180 - 18 = 162

162 = 18g

g = 9

Hence, ∠LMN = (15 x 9 + 18)° = 153°

Therefore, the measure of ∠LMN is 153°.

To learn more about angles, refer to the link:

https://brainly.com/question/14684647

#SPJ1

Answer:

1/4, 0.25

13/40, 0.325

1/25, 0.04

3/4, 0.75

13/200, 0.065

1 1/4, 1.25

1/8, 0.125

1/500, 0.002

3/400, 0.0075

107/100, 1.07

21/10, 2.10

9/40, 0.225

Answer:

value of x i think correct answer is 2

Answer:

1.  1/3

2. 0

Step-by-step explanation:

Answer:

616

Step-by-step explanation:

A fraction that is equivalent to 38 is 616

Answer:

D. point and line

Step-by-step explanation:

Edgunuity

The pair of undefined terms which is used to define a ray is point and line

Option 4 is the correct answer.

Line - It has no fixed points it extends infinitely on both ends.

Line segment - It has two fixed endpoints and does not extend infinitely on any end.

Ray - It has one fixed point on one side and extends infinitely on the other side.

We have to define a ray.

A ray will have a fixed point on one end and extends infinitely on the other end.

i.e a point and a line.

Thus the pair of undefined terms which is used to define a ray is point and line

Learn more about opposite Ray here:

https://brainly.com/question/17491571

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straightAnswer:

Step-by-step explanation:

The strategy would be to look for the equations of  lines that passes for the points. it can be done but it's a hard work. I prefer to use a calc sheet . You can se that house 2 has a perfect fit to data because it has a [tex]R^2=1[/tex], however house 1 does not have a perfect fit, [tex]r^2=0.99..[/tex] altough it is a very good fit, in the image you can see the corresponding equations