5. A car travels 40 kilometers at an average speed of 80 km/h and then travels 40
kilometers at an average speed of 40 km/h. The average speed of the car for this 80 km

trip is:
A) 40 km/h
B) 45 km/h
48 km/h
D) 53 km/h
E) 80 km/h

Answer:

Step-by-step explanation:

It's the square root of the variance

The variance is just the second moment minus the first moment squared

Answer:

i think it is..

Step-by-step explanation:

Answer:

First one would be y=2x (the cost of each ticket is 2 and x is the number of tickets that you are buying)

The second one is y=x+10 (each ticket is one and 10 is the one time set up fee)

9514 1404 393

Answer:

  h = 7

Step-by-step explanation:

Perhaps you want the solution to ...

  1/(h -5) +2/(h +5) = 16/(h^2 -25)

Parentheses are required around denominators that have math operations.

Multiply by (h^2-25) and solve the linear equation.

  [tex]\displaystyle\frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{h^2-25}\qquad\text{given}\\\\(h+5)+2(h-5)=16\qquad\text{multiply by $h^2-25$}\\\\3h-5=16\qquad\text{simplify}\\\\3h=21\qquad\text{add 5}\\\\\boxed{h=7}\qquad\text{divide by 3}[/tex]

Answer:

Discrete Distribution.

Step-by-step explanation:

For each widget, there are only two possible outcomes. Either they develop paint cracks, or they do not. The probability of a widget developing paint cracks is independent of any other widgets, which means that the binomial probability distribution, which is a discrete distribution, is used to solve this question. Thus the answer is a Discrete Distribution.

Answer:

x=161

Step-by-step explanation:

It looks like you have

[tex]f(y)=\dfrac{y^6}{6\ln(y)}-\dfrac{y^6}{36}[/tex]

Differentating gives

[tex]f'(y)=\dfrac{6y^5\times6\ln(y)-6y^6\times\frac1y}{(6\ln(y))^2}-\dfrac{6y^5}{36}= -\dfrac{y^5\left(\ln^2(y)-6\ln(y)+1\right)}{6\ln^2(y)}[/tex]

9514 1404 393

Answer:

Step-by-step explanation:

You know that cos(α) is a maximum at α=0, 2π, 4π, and all even multiples of π. You know cos(α) is a minimum for α=π, 3π, 5π, and all odd multiples of π.

You can find your value of x at which y will be a maximum by setting the argument of the cosine function equal to zero (and/or 2nπ). If we use α=2nπ, then we have ...

  α = (5π/31)(x -3.2) = 2nπ

  (x -3.2) = (31/5)(2n) = 12.4n

Tidal maxima will occur at ...

 x = 12.4n +3.2 . . . . . for integer values of n

Without bothering to go through the solution for α being odd multiples of π, we can see from this that the period is 12.4 hours. We know the tidal minimum  will be half a period later, or 6.2 hours later than this.

Tidal minima will occur at ...

  x = 12.4n +9.4 . . . . for integers n

__

Of course, cos(α) has extremes of ±1, so your tidal maximum will be ...

  y = 11.412 +174.91 = 186.322

and your tidal minimum will be ...

  y = -11.412 +174.91 = 163.498

Answer:

68 degrees

Step-by-step explanation:

Since the angle is a right angle, it is 90 degrees, to figure out the measurement of a section if it, simply subtract the known angle 22, from 90 to get an answer of 68.

Answer:

d = 4.5 cm

A = 1/4 (p x d²)

= 1/4 (3.14 x d x d)

= 1/4 (3.14 x 4.5 cm x 4.5 cm)

= 15.9 cm2

The area of the plate nearest square centimeters is 12cm².

A scalene triangle has three different sides and corresponding to that three different interior angles.

According to the given question we have  triangle with sides  4.5cm,6cm and 7.5cm.

We know for a scalene triangle given 3 sides.

area(A) = [tex]\sqrt{s(s-a)(s-b)(s-c)[/tex].

Where S is semi perimeter and a,b,c are the three sides.

= (a+b+c)/2.

= (4.5+6+7.5)/2 cm.

= 18/2 cm.

= 9 cm.

∴ The area of the triangle is

= [tex]\sqrt{9(9-4.5)(9-6)(9-7.5)[/tex]cm².

= [tex]\sqrt{9(4.5)(3)(1.5)}[/tex] cm².

= [tex]\sqrt{182.5}[/tex] cm² this is in between 13 square and 14 square approx 13.5 cm².

Now it has three small circles of radius of 4 mm or 0.4 cm.

We know area of a circle is πr² and area of 3 circles having same radius is 3(πr²) cm².

= 3{π(0.4)²}

= 3{3.14(0.16)} cm².

= 3(0.5024) cm².

= 1.5072 cm².

Now to obtain the area of the scalene triangle with those three holes of 0.4 cm we have subtract the area of the three circles from the triangle which is

= (13.5 - 1.5) cm².

= 12 cm².

learn more about heron's formula here :

https://brainly.com/question/15188806

#SPJ2

Answer:

19 boys

Step-by-step explanation:

38 kids on the bus

1/2 are girls, that means 1/2 are boys

1/2 * 38 = 19

There are 19 girls and 19 boys

Answer:

about 8/25

Step-by-step explanation:

32.3% = 32/100 = 16/50 = 8/25

rounded percentage down.

put over 100

reduce fraction

Answer:

21; the age of the teacher

Step-by-step explanation:

15+x/2=18

Estimate - I estimated in the 20's because it only averaged up a little bit

Started with 23 and kept going until i got to my answer

21

15+21/2

36/2

18

Please mark brainliest :)

Answer:

Gina would need to take 10 classes.

Step-by-step explanation:

Cost of x classes at dance studio:

[tex]y_d(x) = 15x[/tex]

Cost of x classes at toe tappers:

[tex]y_t(x) = 25 + 12.5x[/tex]

How many classes would Gina need to take for the total cost to be the same at both dance studios?

This is x for which:

[tex]y_d(x) = y_t(x)[/tex]

So

[tex]15x = 25 + 12.5x[/tex]

[tex]2.5x = 25[/tex]

[tex]x = \frac{25}{2.5}[/tex]

[tex]x = 10[/tex]

Gina would need to take 10 classes.

Answer:

10 just took it on edge

Answer:

The difference between the distances they walked is 600 meters.

Step-by-step explanation:

Let's calculate the distance traveled by Nichol and Sakura with the following equation:

[tex] d = v*t [/tex]

Where:

v: is the speed

t: is the time

The distance traveled by Nichol is:

[tex] d_{n} = 1.2 m/s*15min*\frac{60 s}{1 min} = 1080 m [/tex]

And the distance traveled by Sakura is:

[tex] d_{s} = 1.4 m/s*20 min*\frac{60 s}{1 min} = 1680 m [/tex]

Hence, the difference between the distances they walked is:

[tex] d_{t} = d_{s} - d_{n} = 1680 m - 1080 m = 600 m [/tex]

Sakura traveled 600 meters more than Nichol.

Therefore, the difference between the distances they walked is 600 meters.

I hope it helps you!                    

Given:

Number of color pens Tracy have = 63

Number of color pens Jacob have = 46

To find:

How many more colors pens does Tracy have than Jacob​?

Solution:

We need to find the difference between the number of color pens Tracy have and the number of color pens Jacob have.

[tex]Difference=63-46[/tex]

[tex]Difference=17[/tex]

Therefore, Tracy have 17 more color pens than Jacob.

Answer

19.013.153,42629466 giây

Step-by-step explanati

van toc ánh sán  =299.792.458 m/s

s= v*t

t=s/v

t= 5.699.999.999.999.999/299.792.458= 19.013.153,42629466 giây

Answer:

$10.90

Step-by-step explanation:

I don't know why but it wants me to talk when I dont need to

Answer:

Below in bold.

Step-by-step explanation:

We see from the diagram that:

SY = SK + KY

So, substituting the given values:-

36 - x = 13x - 5 + 2x + 9

-x - 13x - 2x = - 5 + 9 - 36

-16x = -32

x = 32/16 = 2.

So SK = 13(2) - 5 = 21.

KY = 2(2) + 9 = 13.

SY = 36 - 2 = 34.

21 + 13 = 34 so this is a check that our calculation is correct.

Answer:

a) 0.0171 fluid ounces.

b) 0% probability that the average fill volume of 22 bags is below 5.95 ounces

c) The mean should be of 6.153 fluid ounces.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Standard deviation of 0.08 fluid ounces.

This means that [tex]\sigma = 0.08[/tex]

(a)What is the standard deviation of the average fill volume of 22 bags?

This is s when n = 22. So

[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

[tex]s = \frac{0.08}{\sqrt{22}}[/tex]

[tex]s = 0.0171[/tex]

(b)The mean fill volume of the machine is 6.16 ounces, what is the probability that the average fill volume of 22 bags is below 5.95 ounces?

We have that [tex]\mu = 6.16[/tex]. The probability is the p-value of Z when X = 5.95. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{5.95 - 6.16}{0.0171}[/tex]

[tex]Z = -12.3[/tex]

[tex]Z = -12.3[/tex] has a p-value of 0.

0% probability that the average fill volume of 22 bags is below 5.95 ounces.

(c)What should the mean fill volume equal in order that the probability that the average of 22 bags is below 6.1 ounces is 0.001?

[tex]X = 6.1[/tex] should mean that Z has a p-value of 0.001, so Z = -3.09. Thus

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]-3.09 = \frac{6.1 - \mu}{0.0171}[/tex]

[tex]6.1 - \mu = -3.09*0.0171[/tex]

[tex]\mu = 6.153[/tex]

The mean should be of 6.153 fluid ounces.

Answer:

a function that converges to 0. '' This means that there is some input size past which the function is always between -0.1 and 0.1; there is some input size past which the function is always between -0.01 and 0.01; and so on.

Answer:

C(min)  =  277.95 $

Container dimensions:

x = 2.822 m

y = 1.411 m

h = 3.52 m

Step-by-step explanation:

Let´s call x  and  y the sides of the rectangular base.

The surface area for  a rectangular container is:

S = Area of the base (A₁) +  2 * area of a lateral side x (A₂) + 2 * area lateral y (A₃)

Area of the base is :

A₁ =  x*y       we assume, according to problem statement that

x  =  2*y      y  = x/2

A₁  =  x²/2

Area lateral on side x

A₂  = x*h           ( h  is the height of the box )

Area lateral on side y

A₃ = y*h        ( h  is the height of the box )

s = x²/2  + 2*x*h + 2*y*h

Cost  =  Cost of the base + cost of area lateral on x + cost of area lateral on y

C = 10*x²/2  + 8* 2*x*h  + 8*2*y*h

C as function of x is:

The volume of the box is:

V(b) = 14 m³  = (x²/2)*h       28 = x²h      h = 28/x²

C(x) = 10*x²/2  + 16*x*28/x² + 16*(x/2)*28/x²

C(x)  =  5*x²  +  448/x  + 224/x

Taking derivatives on both sides of the equation we get:

C´(x)  =  10*x  -  448/x² - 224/x²

C´(x)  =  0             10x  -  448/x² - 224/x² = 0     ( 10*x³ - 448 - 224 )/x² = 0

10*x³ - 448 - 224 = 0       10*x³ = 224

x³  =22.4

x = ∛ 22.4

x = 2.822 m

y = x/2  =  1.411 m

h = 28/x²  =  28 /7.96

h =  3.52 m

To find out if the container of such dimension is the cheapest container we look to the second derivative of C

C´´(x) = 10 + 224*2*x/x⁴

C´´(x)  =  10 + 448/x³    is positive then C has a minimum for x = 2.82

And the cost of the container is:

C = 10*(x²/2) +  16*x*h + 16*y*h

C = 39.82 +  158.75  + 79.38

C =  277.95 $

Answer:

[tex] {x}^{2} + 26x = 11 \\ x = 0.4 \: and \: - 26.4[/tex]

Answer:

-2.5 or - 2 1/2

Step-by-step explanation:

Writing out the expression Mathematically ;

1/3(-15÷1/2)1/4

Using PEMDAS :

Solving the bracket first

(-15 ÷ 1/2) = (-15 * 2/1) = - 30

We have :

1/3(-30)1/4 = - 10 * 1/4 = - 10 / 4 = - 2.5

-2.5 = - 2 1/2

Answer: 57%

Step-by-step explanation:

Answer:

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

Step-by-step explanation:

The area for a rectangle is given by the formula:

[tex]A=w\ell[/tex]

Where w is the width and l is the length.

We are given that the length of the rectangle is increasing at a rate of 6 cm/s and that the width is increasing at a rate of 5 cm/s. In other words, dl/dt = 6 and dw/dt = 5.

First, differentiate the equation with respect to t, where w and l are both functions of t:

[tex]\displaystyle \frac{dA}{dt}=\frac{d}{dt}\left[w\ell][/tex]

By the Product Rule:

[tex]\displaystyle \frac{dA}{dt}=\frac{dw}{dt}\ell +\frac{d\ell}{dt}w[/tex]

Since we know that dl/dt = 6 and that dw/dt = 5:

[tex]\displaystyle \frac{dA}{dt}=5\ell + 6w[/tex]

We want to find the rate at which the area is increasing when the length is 12 cm and the width is 4 cm. Substitute:

[tex]\displaystyle \frac{dA}{dt}=5(12)+6(4)=84\text{ cm}^2\text{/s}[/tex]

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

Answer:

11j - 1573/10

Step-by-step explanation:

J = 10v /11 +143

collect terms

J - 143 =10v /11

11 (J -143) = 10v

Divide both sides by 10

11(J - 143)/10 = v

v = 11J - 1573/10

Hope this helps!