CAN SOMEONE PLEASE HELL ME WITH THIS PROBLEM? THANK YOU!!!

Answer:

5

Step-by-step explanation:

5+7

9514 1404 393

Answer:

Step-by-step explanation:

The range of the function is the vertical extent. Here, the values of y can be anything that is -2 or more:

  range: -2 ≤ y

The domain of a the function is the horizontal extent. The domain of any polynomial function is ...

  domain: All reals

Answer:

Sports is beneficial to physical health and well-being

as well as for growth and development, it decreases chances of contracting diseases and illnesses such as cardiovascular disease, lowers fat - intake as well as improves memory and relieves stress.

Answer:

6x+y

Step-by-step explanation:

Answer:

Square Root of 529:√529 = 23

Square Root of 1080: 32.863353450309965

Answer:

Step-by-step explanation:

First, I would find the point on the curve. By substituting t=1, I get (x, y). Next, I will try to eliminate the t and make a xy equation. In this case, the t's will cancel out in 'x=t-t-1" which wouldnt make this a curve. To find the equation of the tangent line, find the deretitave of the xy equation, and subsitute x in to find the slope at that point. Next, use point slope form to find the equation at the point.

Answer:

There are 45 dogs, 27 cats

Step-by-step explanation:

Write and solve an equation of ratios.  The "cat ratio" is 3/27 and the "dog ratio" is 5/d:

Then

 3         5

------ = -----

 27       d

Through cross-multiplication, we get 3d = 135, or d = 45

There are 45 dogs, 27 cats.  Notice that 27/45 = 3/5, as expected.

There are 45 dogs and 27 cats.  

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Given information;The ratio of cats to dogs at the animal shelter is 3 to 5.

There are 27 cats.

We need to Write and solve an equation of ratios.  

The "cat ratio" is 3/27 and the "dog ratio" is 5/d:

Then 3/ 27 =   5 / d

Through cross-multiplication, we get

d = 5 x 27 / 3d = 45So,

Notice that 27/45 = 3/5, as expected.

There are 45 dogs and 27 cats.  

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Answer:

Step-by-step explanation:

As,

[tex] \frac{2}{3} = \frac{2 \times 10}{3 \times 10} = \frac{20}{30} [/tex]

and,

[tex] \frac{3}{10} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30} [/tex]

Answer:

Domain -1 ≤x<2

Range 0 < y ≤4

Step-by-step explanation:

Domain is the input values

X goes from -1 to 2 ( 2 not included)

Domain -1 ≤x<2

Range is the output values

y goes from 0 ( not included) to 4

Range 0 < y ≤4

Answer:

Since [tex]n(1-p) = 3.6 < 5[/tex], the normal distribution cannot be used as an approximation to the binomial probability to approximate the probability.

Using the binomial distribution, 100% probability that more than 97 out of 120 people will get the flu this winter.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

Assume the probability that a given person will get the flu this winter is 97%.

This means that [tex]p = 0.97[/tex]

120 people

This means that [tex]n = 120[/tex]

Verifying the necessary conditions.

[tex]np = 120*0.97 = 116.4[/tex]

[tex]n(1-p) = 120*0.03 = 3.6[/tex]

Since [tex]n(1-p) = 3.6 < 5[/tex], the normal distribution cannot be used as an approximation to the binomial probability to approximate the probability. Thus, the binomial distribution has to be used.

Probability using the binomial distribution:

[tex]P(X > 97) = P(X = 98) + ... + P(X = 120)[/tex]

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

[tex]P(X = 98) = C_{120,98}.(0.97)^{98}.(0.03)^{22} \approx 0[/tex]

Probability close to 0, but below the mean, which means that the probability of the number being above this is 100%.

Using the binomial distribution, 100% probability that more than 97 out of 120 people will get the flu this winter.

Answer

You need to pay $14.20 to get 4 coffees and 6 donuts.

Explanation

Let's say that the price of one coffee and x and the price of one donut is y.

In the first instance, 3x+4y=10.05.

In the second instance, 5x+7y=17.15.

You can use these equations to find the value of a coffee and the value of a donut.

We can find x and y using elimination. To do this, you should add or subtract one equation from another to "get rid" of a variable (I'll "get rid" of x). We can't just add or subtract the equations right now, since that wouldn't lead to 0x.

Multiply the first equation by 5, and the second equation by 3. After this, both equations will have 15x. Make sure to multiply each term by 5 and 3.

15x+20y=50.25. This means that 15 coffees and 20 donuts is $50.25.

15x+21y=51.45. This means that 15 coffees and 21 donuts is 51.45.

Now since there are an equal number of coffees, we can subtract these equations.

(15x+21y=51.45)-(15x+20y=50.25) equals y=1.20 (a donut costs $1.20).

Plug the price of the donut into y to find x; 3x+4(1.20)=10.05. The value of x is 1.75. The price of a coffee is 1.75.

You can multiply 1.75 by 4 to find the price of 4 coffees and 1.20 by 6 to find the price of 6 donuts.

1.75*4 is 7.00 and 1.20*6 is 7.20.

You can add those to find the total price; 7.00+7.20=14.20.

Answer: A=πr²      

A=3.14(1.6inch)²                     r=d/2⇒3.2/2⇒1.6

A=3.14×2.56in²

A=8.0384in²

A≈8.04

now circumference,

C=2πr

C=2×3.14×1.6in

C=10.048in

C≈10.05

Answer:

Step-by-step explanation:

Probability(P) (k events out of n trials) = nCk * p^k * (1-p)^(n-k), where p=0.40, n=10 and nCk is the number of combinations of n things taken k at a time

:

P ( k < or = 3 ) = P ( k = 3 ) + P ( k = 2 ) + P ( k = 1 ) + P ( k = 0 )

:

P ( k = 3 ) = 10C3 * (0.40)^3 * (0.60)^(10-3) = 0.2149

:

P ( k = 2 ) = 10C2 * (0.40)^2 * (0.60)^(10-2) = 0.1209  

:

P ( k = 1 ) = 10C1 * (0.40)^1 * (0.60)^(10-1) = 0.0403

:

P ( k = 0 ) = 10C0 * (0.40)^0 * (0.60)^(10-0) = 0.0060

:

******************************************************************************

P ( k < or = 3 ) = 0.2149 + 0.1209 + 0.0403 + 0.0060 = 0.3821 is approximately 0.38

The probability that when 14 adults are randomly selected and 3 or fewer are in excellent health is 0.072

Probability shows possibility to happen an event, it defines that an event will occur or not. The probability varies from 0 to 1.

Given that,

The percentage of adults who reported their health was excellent = 44 %

The probability of excellent health P = 0.44

The probability of not excellent health = Q - 1 - 0.44 - 0.56

Since, 14 adults are randomly selected.

To find the probability that 3 or fewer are in excellent health,

Use formula

Probability = [tex]^nC_rP^n Q^{n-r}[/tex]

Where P = probability to happen an event, and Q= probability to not happen an event.

The probability that 3 or fewer are in excellent health

= [tex]^{14}C_3 (0.44)^3(0.56)^{14-3} +^{14}C_2(0.44)^2(0.56)^{14-2}+^{14}C_1 (0.44)^1(0.56)^{14-1} + ^{14}C_0 (0.44)^0(0.56)^{14}[/tex]

= 0.0526 + 0.0167 + 0.00328 + 0.000298

= 0.072282

The required probability is 0.072282.

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X = (18/5)/(3/2)

= 18/5 x 2/3

=(18 x 2) / (5x3)

= 90/15

If you don’t want it as a fraction it’s 6!

Hope this helps!

Answer:

yeah, teachers kinda suck

In polar coordinate, R is the set of points

{(r, θ) | 1 < r < 5 and 0 < θ < π/2}

So the integral is

[tex]\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA = \int_0^{\frac\pi2}\int_1^5 r\sin(r^2)\,\mathrm dr\,\mathrm d\theta[/tex]

[tex]=\displaystyle\frac\pi2\int_1^5 r\sin(r^2)\,\mathrm dr[/tex]

[tex]=\displaystyle\frac\pi4\int_1^5 2r\sin(r^2)\,\mathrm dr[/tex]

[tex]=\displaystyle\frac\pi4\int_1^{25} \sin(s)\,\mathrm ds[/tex]

(where s = r ²)

[tex]=\displaystyle-\frac\pi4\cos(s)\bigg|_1^{25}= \boxed{\dfrac\pi4 (\cos(1) - \cos(25))}[/tex]

Answer:

1/6

Step-by-step explanation:

We want to find how long it takes for the bacteria to lose half its size. We can do this by taking one point of the bacteria and finding how long it takes to go to half its size. When t=0, 9300 * (1/64)^t = 9300 * 1 = 9300 as anything to the power of 0 is 1. Therefore, we can solve for t when the end result of the bacteria is 9300/2= 4650, making our equation

4650 = 9300 * (1/64)^t

divide both sides by 9300

1/2 = (1/64)^t

First, we can tell that 2^6 = 64*. Because of this, we can say that (1/2)^6 = 1^6/2^6 = 1/64, so (1/64)^(1/6) = 1/2. We know this because

(1/2)^6 = 64

take the 6th root of both sides

(1/2) = (64)^(1/6)

. This means that t=1/6, so the bacterial culture loses 1/2 of its size every 1/6 seconds

* if this is harder to figure out, e.g. 3 and 729, we can plug (log₃729) into a calculator

Answer:

0.17 seconds

Step-by-step explanation:

i got this correct on Khan   :)

i hope it helps

Answer:

a: 1/12

b: 1/6

c: 1/2

d: 1/2

e: 1/12

f: 1/3

Step-by-step explanation:

Answer:

3 =f

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 60 = f/ sqrt(3)

sqrt(3) tan 60 = f

sqrt(3) * sqrt(3) = f

3 =f

9514 1404 393

Answer:

  162 mi²

Step-by-step explanation:

The area on the map is ...

  18(1 in)²

Then the area on the ground will be ...

  18(3 mi)² = 18·9 mi² = 162 mi²

Answer:

The probability of both dice having the same number is 636, as there are 36 different outcomes, 6 of which have two of the same number, i.e. (1,1),(2,2),....

The expected number of rolls of this type in 100 pairs of dice rolls is 100∗636

Answer:

4:16 = 0.25

Step-by-step explanation:

16 outcomes are possible, 4 with the same number.

9514 1404 393

Answer:

  3 ⇒ 12

Step-by-step explanation:

Apparently "a = b" in this case is used to mean f(a) = b. It appears as though the function is ...

  f(x) = x(x+1)

Then f(3) = 3(3+1) = 3·4 = 12

_____

Additional comment

IMO this is a poor use of the equal sign, which should be reserved for situations where the left side expression has the same value as the right side expression.

Answer:

BF = 16

Step-by-step explanation:

18/12 = 1.5 * 6 = 9

Since DE and BF are parallel and DB and EF are parallel, they comprise a parallelogram. This means that DB = EF

DB = EF = 9

24/1.5 = 16

DE = 16

BF = 16

The statement which can be proven true using the given theorem (congruence) is Segment BF = 16.

By the congruence theorem;

Hence, AE/EC = BF/FC.

Therefore; 12/18 = BF/24

Hence, BF = 24× 12/18

Read more on congruent triangles;

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Answer:

x=71

Step-by-step explanation:

Since this is an isosceles triangle as indicated by the lines on the sides, the sides lengths are equal.

When the sides are equal, the base angles are equal

x=71

9514 1404 393

Answer:

  (b)  AB overbar congruent to CD overbar

Step-by-step explanation:

AB with an overbar is the way that line segment AB is designated, where appropriate typesetting is possible. Thus the statement the line segments are congruent is fully equivalent to ...

  AB overbar congruent to CD overbar

Answer:

The equation will be "[tex]y=x-3[/tex]".

Step-by-step explanation:

Given:

Points (12, 9) = (x, y)

⇒ [tex]x=9t^2+3[/tex]

then,

 [tex]\frac{dy}{dt}=18t[/tex]

or,

⇒ [tex]y=6t^3+3[/tex]

then,

 [tex]\frac{dy}{dt}=18t^2[/tex]

⇒ [tex]\frac{dy}{dx}=\frac{18t^2}{18t}[/tex]

        [tex]=t[/tex]

By using the point slope form.

The equation of tangent will be:

⇒ [tex]y-9=1(x-12)[/tex]

   [tex]y-9=x-12[/tex]

          [tex]y=x-12+9[/tex]

          [tex]y=x-3[/tex]

Answer:

6

Step-by-step explanation:

The given function to us is ,

[tex]\rm\implies f(x)= log_a(x) [/tex]

And its value at 7 is 2 , that is ,

[tex]\rm\implies f(x)= log_a(7) =2[/tex]

Taking this ,

[tex]\rm\implies 2= log_a(7) [/tex]

In general we know that ,

[tex]\bf\to log_a b = c ,\ then \ a^c = b [/tex]

Using this , we have ,

[tex]\rm\implies a^2 = 7 [/tex]

Squarerooting both sides ,

[tex]\rm\implies a =\sqrt{ 7 }[/tex]

Therefore , when x is 343 ,

[tex]\rm\implies f(343)= log_{\sqrt7} ( 343) [/tex]

We can write , 343 as 7³ ,

[tex]\rm\implies f(343)= log_{\sqrt7}7^3 [/tex]

[tex]\rm\implies f(343)= log_{7^{\frac{1}{2}}} 7^3 [/tex]

This can be written as ,

[tex]\rm\implies f(343)= \dfrac{ 3}{\frac{1}{2}} [/tex]

[tex]\rm\implies \boxed{\blue{\rm f(343)= 6 }}[/tex]

Hence the required answer is 6.

Answer:

tk 30

Step-by-step explanation:

5 dozen = 5 * 12 = 60 oranges

12/4 = 3

total cost = 60 * 3 = tk.180

total sell = 50 * 3 = tk 150

total lose = 180 - 150 = tk 30

Answer:

18

Step-by-step explanation:

Shape = Isosceles Trapezoid

Area = 1/2 (a+b) h

= 1/2 (9+3) x 3

= 18

Answered by Gauthmath

Answer:

D (I think?)

Step-by-step explanation:

It's neither a nor b because neither of them would work if you plug in o for x.

The reason I think it's D and not C is because it goes up by 10,000 x (1.1)^x

Just not sure because because of how the question was asked. Hope this helps

Answer:

attributed to enhanced NAVs made possible by brands’ disclosure in the Balance

Sheet. However, there is a biggest handicap of reliable and independent assessment

of the brand value. This is the single most important reason for the brands not

allowed to be placed on the Balance Sheet by the Authorities concerned. Though,

value is assigned to externally acquired brands, there is no effective way for

assessment of home-grown brands. Nevertheless, it is gratifying to note that the

Institute of Chartered Accountants of England and Wales (ICAE & W), has given due

recognition to the internally generated brand values and allow disclosure in the

Balance Sheet which necessitates periodic revaluation.

The brands are to be explicitly recognized, as part of purchased Goodwill, by the

ICAI of India, which would make possible the recognition of brand values. Rules and

methodologies necessary for the valuation accounting and disclosure of brands may

be framed in this regard. This would go a long way in governing the brand

accounting practice and contribute to healthy business combinations in India,

especially where the brand values are given serious consideration in the emerging

strategic alliances, mergers and acquisitions fueled by the entry of MNCs. This

assumes special significance in the environment of globalization, liberalization and

privatization of Indian economy.

It emerges from the above discussions that Goodwill, brand valuation and

accounting are definite to take the world of accounting by storm in the near future

which would mark the beginning of a new era of accounting practice. Thus, there

exists a strong and a clear case for brand accounting, especially in view of the

imminent benefits of brand accounting to the corporate bodies and the investors alike.

15.13 Self Assessment Questions

1. Define Brand. Discuss the objectives of corporate branding ?

2. What is the necessity of brand accounting in the competitive business environment?

3. Discuss the difficulties in accounting the brands.

4. Explain various models available for valuation of home-grown brands.

5. How would you value the acquired brands ?

15.15 SUGGESTED READINGS

1. Kotler Philip, Armstrong Gary, Principles of Marketing, Printice-Hall of India (Pvt.)

Ltd.

2. ASB, Goodwill and Intangible Asset, U.K.

3. Pyne Radhanath, Valuing brands, The Accounting Debate, The Chartered Accountant.

4. Datta Manipadma, Brand Equity : A paradigm shift in firm valuation, Chartered

Secretary.