Please answer this in two minutes

Answer:

x+y = 14

Step-by-step explanation:

x+y+xy = 54

1+x+y+xy = 55

(1+x)(1+y) = 55

(1+x)(1+y) = 5*11

If x and y are integers,

1+x = 5    => x = 4

1+y = 11   =>  y = 10

or

1+x = 11   => x = 10

1+y = 5   =>  y = 4

In either case, x+y = 14

The solution to this system is [tex]x = 4[/tex] and [tex]y = 10[/tex]. Then, [tex]x+y = 14[/tex].

Let be [tex]x + y + x\cdot y = 54[/tex], then we proceed to modify the expression by algebraic means:

The only divisors of 55 is 5 and 11, respectively. Hence, we have the following  system of linear equations:

[tex]x + 1 = 5[/tex] (1)

[tex]y + 1 = 11[/tex] (2)

The solution to this system is [tex]x = 4[/tex] and [tex]y = 10[/tex]. Then, [tex]x+y = 14[/tex].

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Answer:

9120(b)

Step-by-step explanation:

95×96

(90+5)×(90+6)

(90×90)+(5+6) (90)+(5)(6)

(90) square+(5+6) (90)+(5)(6)

I am using identity 4

=8100+990+30=9120

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = the interval of time between the eruptions

So, X ~ N([tex]\mu=72, \sigma^{2} =23^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                            Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean time = 72 minutes

           [tex]\sigma[/tex] = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{82-72}{23}[/tex] ) = P(Z > 0.43) = 1 - P(Z [tex]\leq[/tex] 0.43)

                                                           = 1 - 0.6664 = 0.3336

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let [tex]\bar X[/tex] = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

                            Z  =  [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean time = 72 minutes

           [tex]\sigma[/tex] = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P([tex]\bar X[/tex] > 82 min)

       P([tex]\bar X[/tex] > 82 min) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{82-72}{\frac{23}{\sqrt{13} } }[/tex] ) = P(Z > 1.57) = 1 - P(Z [tex]\leq[/tex] 1.57)

                                                           = 1 - 0.9418 = 0.0582

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let [tex]\bar X[/tex] = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

                            Z  =  [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean time = 72 minutes

           [tex]\sigma[/tex] = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P([tex]\bar X[/tex] > 82 min)

       P([tex]\bar X[/tex] > 82 min) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{82-72}{\frac{23}{\sqrt{34} } }[/tex] ) = P(Z > 2.54) = 1 - P(Z [tex]\leq[/tex] 2.54)

                                                           = 1 - 0.9945 = 0.0055

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

Answer:

Step-by-step explanation:

From the given data

mean, u = 72

Standard deviation [tex]\rho[/tex] = 23

A) Probability that a randomly selected time interval between eruptions is longer than 82​minutes

[tex]P (x > 82) = P[\frac{x-u}{\rho} > \frac{82-72}{23}]\\\\P (x > 82) = P[z > 0.43]\\\\P (x > 82) = 0.3336[/tex]

B)

[tex]P (x > 82) = P[\frac{x-u}{\frac{\rho}{\sqrtn}} > \frac{82-72}{\frac{23}{\sqrt{13}}}]\\\\P (x > 82) = P[z > 1.5676]\\\\P (x > 82) = 0.0594[/tex]

C)

[tex]P (x > 82) = P[\frac{x-u}{\frac{\rho}{\sqrtn}} > \frac{82-72}{\frac{23}{\sqrt{34}}}]\\\\P (x > 82) = P[z > 2.5351]\\\\P (x > 82) = 0.0057\\\\[/tex]

D) If the mean is less than 82minutes, then the probability that the sample mean of the time between eruptions is greater than 83 minutes decrease because the variability in the sample mean decrease as the sample size increases

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Answer:

12-5=7

unless it's not a prank or a joke question

Answer:

7

Step-by-step explanation:

Original number of monkeys = 12

Number taken away = 5

So, number left = 12-5 = 7.

Hope this helps.

Answer:

D. 4√2

Step-by-step explanation:

A triangle with 45°, 45°, and 90° is a special right triangle.

hypotenuse = √2 · leg

1. Set up the equation

8 = √2 · x

2. Divide by √2 and solve

x = [tex]\frac{8}{\sqrt{2} }[/tex] · [tex]\frac{\sqrt{2} }{\sqrt{2}}[/tex] = [tex]\frac{8\sqrt{2} }{2}[/tex] = 4√2

Answer:

x = 4.2

Step-by-step explanation:

To calculate for the value of x, we must first calculate the value of y as shown in the attached photo.

The value of y can be obtained as follow:

Angle θ = 30°

Opposite = y

Hypothenus = 6

Sine θ = Opposite /Hypothenus

Sine 30° = y/6

Cross multiply

y = 6 × Sine 30°

y = 3

Therefore, the value of y is 3.

Now, we can obtain the value of x as follow:

Angle θ = 45°

Adjacent = y = 3

Hypothenus = x

Cos θ = Adjacent /Hypothenus

Cos 45° = 3/x

Cross multiply

x × Cos 45° = 3

Divide both side by Cos 45°

x = 3/Cos 45°

x = 4.2

Therefore, the value of x is 4.2

Answer:

x₁ = (1 + √43)/6

x₂ = (1 - √43)/6

Step-by-step explanation:

6x² - 2x = 7

6x² - 2x - 7 = 0

Formula

ax² + bx + c = 0

a = 6 ; b = - 2 ; c = - 7

x₁/₂ = [-b ± √(b²- 4ac)]/2a

x₁/₂ = (2 ± √4 + 4×6×-7)/12

x₁/₂ = (2 ± √172)/12

x₁/₂ = (2 ± 2√43)/12

x₁/₂ = 2(1 ± √43)/12

x₁/₂ = (1 ± √43)/6

x₁ = (1 + √43)/6

x₂ = (1 - √43)/6

Answer:

Both the relations are functions, the correct answer is a.

Step-by-step explanation:

In order to solve this problem we will first find the inverse relation as shown below:

[tex]y = 3x^2 + 5\\x = 3y^2 + 5\\3y^2 = x - 5\\y^2 = \frac{x - 5}{3}\\y = \sqrt{\frac{x - 5}{3}} = \frac{\sqrt{x - 5}}{\sqrt{3}}\\y = \frac{\sqrt{x - 5}\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\sqrt{3x - 15}}{3}[/tex]

Functions are relations between two groups of numbers, for which the input must generate only one output. Using this definition we can classify both the relation and its inverse as a function, therefore the correct answer is a.

Answer:

[tex]\large \boxed{\sf \ \ \ \dfrac{\sqrt{4025}-5}{2}=29.22144... \ \ \ }[/tex]

Step-by-step explanation:

Hello

Let's note the original speed of the car v

it means that in 1 hour he is going v miles

so to go 200 miles it takes ( in hour)

[tex]\dfrac{200}{v}[/tex]

If the speed of the car is v+5 than to go 200 miles it takes (in hour)

[tex]\dfrac{200}{v+5}[/tex]

and this time is one hour less so we can write

[tex]\boxed{\sf \ \ \dfrac{200}{v+5}=\dfrac{200}{v}-1 \ \ }[/tex]

We can multiply by v(v+5) both parts of the equation so

[tex]200v=200(v+5)-v(v+5)\\\\<=>200v=200v+1000-v^2-5v\\\\<=>v^2+5v-1000=0[/tex]

[tex]\Delta=b^2-4ac=5^2+4*1000=4025\\\\ \text{There are potential solutions }\\\\\ \ \ \ \ x_1=\dfrac{-5-\sqrt{4025}}{2}\\\\\ \ \ \ \ x_2=\dfrac{-5+\sqrt{4025}}{2}[/tex]

Only one is positive and this is is

[tex]x_1=\dfrac{\sqrt{4025}-5}{2}=29.22144...[/tex]

So the original speed is 29.22144... mph

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

2267676767/1000000000

Step-by-step explanation:

To write 2.267676767 as a fraction you have to write 2.267676767 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

2.267676767 = 2.267676767/1 = 22.67676767/10 = 226.7676767/100 = 2267.676767/1000 = 22676.76767/10000 = 226767.6767/100000 = 2267676.767/1000000 = 22676767.67/10000000 = 226767676.7/100000000 = 2267676767/1000000000

And finally we have:

2.267676767 as a fraction equals 2267676767/1000000000

Hope it helps.. If yes mark me BRAINLIEST!!

TYSM!

Answer:

2267676767/1000000000

Step-by-step explanation:

when converting a decimal to fraction,

write down the decimal divided by 1 and than add as many zeros as the digits after the decimal point.

since we have nine digits after the decimal point  nine zeros are put after the '1'

Answer: √3

Step-by-step explanation:

Hi, since the situation forms a right triangle we have to apply the next trigonometric function.  

Sin α = opposite side / hypotenuse  

Where α is the angle (30°), the hypotenuse is the longest side of the triangle (in this case 2), and the opposite side is x. (not t or 2)

Replacing with the values given:  

Sin 30 = x /2

Sin 30 (2) = x

x=1

Applying the Pythagorean theorem:

c^2 = a^2 + b^2  

Where c is the hypotenuse of the triangle (2) and a and b are the other sides.  

2^2 = t^2 + 1^2

4= t^2 +1

4-1 = t^2

3 = t^2

√3 = t

Feel free to ask for more if needed or if you did not understand something.  

Answer:

>There is a correlation between number of bottles bought and number of samples offered. However, there is no causation. This is because there is probably an increase in the number of bottles bought with an increase in the number of samples offered.

>There is no correlation between playing time and number of goals.

Hope this helps....

Have a nice day!!!!

Answer;

The relevant probability is 0.136 so the value of 56 girls in 100 births is not a significantly high number of girls because the relevant probability is greater than 0.05

Step-by-step explanation:

The complete question is as follows;

For 100 births, P(exactly 56 girls = 0.0390 and P 56 or more girls = 0.136. Is 56 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question? Consider a number of girls to be significantly high if the appropriate probability is 0.05 or less V so 56 girls in 100 birthsa significantly high number of girls because the relevant probability is The relevant probability is 0.05

Solution is as follows;

Here. we want to know which of the probabilities is relevant to answering the question and also if 56 out of a total of 100 is sufficient enough to provide answer to the question.

Now, to answer this question, it would be best to reach a conclusion or let’s say draw a conclusion from the given information.

The relevant probability is 0.136 so the value of 56 girls in 100 births is not a significantly high number of girls because the relevant probability is greater than 0.05

Answer:

a. 7.5 minutes

b. 4 and one-half toothpicks

Step-by-step explanation:

a. an hour = 60 minutes

Allotted time for the exam = 1/2 hour = 1/2 * 60 = 30 minutes

He used 3/4 of this time, so the time left in fraction would be 1-3/4 = 1/4

So the time left in minutes is 1/4 * 30 = 7.5 minutes

b. The total length he has here is 14 3/5 inches

Now he wants to cut toothpicks of 3 1/5

To know the number of toothpicks he can cut, we simply divide the total length by the length of each toothpick

14 3/5 divided by 3 1/5

= 73/5 divided by 16/5

= 73/5 * 5/16 = 73/16 = 4.5625

which is closer to 4 and one-half toothpicks

Answer:

Step-by-step explanation:

Let's solve your inequality step-by-step.

(-34)(x)≤12

Step 1: Simplify both sides of the inequality.

−34x≤12

Step 2: Multiply both sides by 4/(-3).

(4−3)*(−34x)≤(4−3)*(12)

x≥−16

Answer:

x≥−16

IM A MATH GOD

Answer:

The dimensions of the rectangular pen that maximize the area are [tex]w = 125\,ft[/tex] and [tex]l = 250\,ft[/tex].

Step-by-step explanation:

Let suppose that one side of the rectangular area to be fence coincides with the contour of the river, so that only three sides are needed to be enclosed. The equations of perimeter ([tex]p[/tex]) and area ([tex]A[/tex]), measured in feet and square feet, are introduced below:

[tex]p = 2\cdot w + l[/tex]

[tex]A = w\cdot l[/tex]

Where [tex]w[/tex] and [tex]l[/tex] are the length and width of the rectangle, measured in feet.

Besides, let suppose that perimeter is equal to the given amount of fencing, that is, [tex]p = 500\,ft[/tex]. The system of equations is:

[tex]2\cdot w + l = 500\,ft[/tex]

[tex]A = w\cdot l[/tex]

Let is clear the length of the rectangle and expand the area formula:

[tex]l = 500\,ft-2\cdot w[/tex]

[tex]A = w\cdot (500\,ft-2\cdot w)[/tex]

[tex]A = 500\cdot w -2\cdot w^{2}[/tex]

To determine the maximum area that can be enclosed, first and second derivatives to obtain the critical values that follow to an absolute maximum.

First derivative

[tex]A' = 500 - 4\cdot w[/tex]

Second derivative

[tex]A'' = -4[/tex]

Now, let equalize the first derivative to zero, the only critical value is:

[tex]500-4\cdot w = 0[/tex]

[tex]4\cdot w = 500[/tex]

[tex]w = 125\,ft[/tex]

Since the second derivative is a negative constant function, then, the previous outcome follows to an absolute maximum. The length of the rectangular area is: ([tex]w = 125\,ft[/tex])

[tex]l = 500\,ft - 2\cdot (125\,ft)[/tex]

[tex]l = 250\,ft[/tex]

The dimensions of the rectangular pen that maximize the area are [tex]w = 125\,ft[/tex] and [tex]l = 250\,ft[/tex].

Answer:

(-2,-3)

Step-by-step explanation:

abscissa is always know as the x axis

ordinate is known as y axis

therefore

it lies in the fourth quadrant

Answer:

Length of the arc = π/50 cm

Step-by-step explanation:

Hello,

The diagram above is a circle and the to find the length of the arc, we need to know the formula for that.

Length of an arc = θ/360 ×2πr

θ = 3π/5

radius (r) = 6cm

Substitute the values into the equation

[(3π/5) / 360] × 2πr

Length of an arc = 3π/5 × ¹/₃₆₀ × 2π(6)

Length of an arc = (3π x 12π) / (5×360)

Length of an arc = 36π/1800

Length of an arc = π/50cm

The length of the arc is π/50cm

Answer:

last answer

Step-by-step explanation:

P' (2, -4)

Q' (-2, -5)

R' (1, -8)

Answer:

C. P'(2, -4) Q'(-2, -5) R'(1, -8)

Step-by-step explanation:

When you reflect something across the y-axis you change (x,y) to (-x,y).

For each point, change the x to a negative x.

P(-2, -4) --> P'(2, -4)

Q(2, -5) --> Q'(-2, -5)

R(-1, -8) --> R'(1, -8)

Hope this helps. If you have any follow-up questions, feel free to ask.

Have a great day! :)

Answer:

Option D. f(x) = 6^x

Step-by-step explanation:

To know which of the function is increasing, let us obtain f(1) and f(2) for each function.

This is illustrated below:

f(x) = (1/6)^x

f(1) = (1/6)¹ = 1/6

f(2) = (1/6)² = 1/36

Therefore, f(x) = (1/6)^x is decreasing.

f(x) = (0.6)^x

f(1) = (0.6)¹ = 0.6

f(2) = (0.6)² = 0.36

Therefore, f(x) = (0.6)^x is decreasing.

f(x) = (1/60)^x

f(1) = (1/60)¹ = 1/60

f(2) = (1/60)² = 1/3600

Therefore, f(x) = (1/60)^x is decreasing.

f(x) = 6^x

f(1) = 6¹ = 6

f(2) = 6² = 36

Therefore, f(x) = 6^x is increasing.

Answer:

Option D

Step-by-step explanation:

The reason why it is D is because if it was something below 1, such as 0.6, it would be decreasing. That is why 6 is the answer.

Answer:

Part A Part B Part C explained

Step-by-step explanation:

PART A: Extrema: relative minimums (4,-2) and maybe (7,-3) (uncertain because it’s an endpoint), relative maximums (6,0) and maybe (1,6) (uncertain because it’s an endpoint)

Zeros: (6,0), somewhere between t = 3 and t = 4 since the graph changes from positive to negative, and somewhere possibly between t = 6 and t = 7 unless the graph hits (6,0) and stays negative.

End behavior: We can guess that as x approaches infinity, the functions approaches negative infinity, and as x approaches negative infinity, the function approaches infinity.

Intervals of increase and decrease: Increasing on (4, 6), decreasing on (1, 4) and (6, 7)

PART B: The relative minimums indicate that the two lowest temperatures occurred on day 4 at -2°F and day 7 at -3°F. The relative maximums indicate that the weekly highs were day 1 at 6°F and day 6 at 0°F.

The zeros of the function represent when the temperature in Johnstown was 0°F. This happened sometime between days 3 and 4 and on day 6.

In the context of the problem, it doesn’t make sense to go an infinite number of degrees below zero. And, the end behavior is ignored because of the restricted range.

The intervals of increase indicate when the temperature is rising, and the intervals of decrease indicate when the temperature is dropping. The intervals where the values are positive indicate when the temperature is above 0°F. The intervals where the values are negative indicate when the temperature is below 0°F.

PART C: The domain is restricted to the number of days the town recorded the temperature. So, the domain is [1, 7].

The range represents the range of temperatures of Johnstown over the course of one week. So, the range is [-3, 6].

Answer:

Part A was missing the last so this is the correct answer.

Extrema: relative minimums (4,-2) and maybe (7,-3) (uncertain because it’s an endpoint), relative maximums (6,0) and maybe (1,6) (uncertain because it’s an endpoint)

Zeros: (6,0), somewhere between t = 3 and t = 4 since the graph changes from positive to negative, and somewhere possibly between t = 6 and t = 7 unless the graph hits (6,0) and stays negative.

End behavior: We can guess that as x approaches infinity, the functions approach negative infinity, and as x approaches negative infinity, the function approaches infinity.

Intervals of increase and decrease: Increasing on (4, 6), decreasing on (1, 4) and (6, 7)

Positive and negative intervals: Positive from 1 to somewhere between t = 3 and t = 4, and negative from somewhere between t = 3 and t = 4 to t = 6, and from some point after t = 6 to t = 7.

But other than that everything else was correct and thank you.

Step-by-step explanation:

Answer:

-2

Step-by-step explanation:

3n+2n+4=9n+12

5n+4=9n+12

5n-9n=-4+12

-4n=8

n=8/-4

n= -2

Answer:

-2

Step-by-step explanation:

I got this right in my assignment.

Answer:

The correct option is;

Choice A 4·(n - 3) - 6·n

Step-by-step explanation:

The given expression is

Which gives;-6·n+(-12)+4·n

- 12 + 4·n-6·n = -2·n - 12 = - (2·n + 12)

The options given are Choice A and/or Choice B;

(Choice A) 4·(n - 3) - 6·n

Which can be simplified as follows;

4·(n - 3) - 6·n = 4·n - 12 - 6·n

Which gives;

4·n - 12 - 6·n = 4·n  - 6·n- 12 = -2·n - 12 = -(2·n + 12)

Therefore, 4·(n - 3) - 6·n is equivalent to -6·n+(-12)+4·n

For choice B, we have;

2·(2·n - 6) which gives;

2·(2·n - 6) = 4·n - 12

Therefore, 2·(2·n - 6) is not equivalent to -6·n+(-12)+4·n

Which gives the correct option as Choice A.

4(n-3)-6n

Khan academy I got this right

Answer:3

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

[tex](-a+3)(a+4)=-a^2-a+12[/tex].

Hope this helps.

Answer:

-a²-a+12

Step-by-step explanation:

-a²+3a-4a+12

-a²-a+12

Answer: B.) $275

Step-by-step explanation:

Insurance cost :

Housing insurance - - - 50c per $100

Content insurance - - - 25c per $100

House valuation = $50,000

50c = $0.5

Therefore insurance cost on House:

Valuation × insurance cost

($50,000 / $100) * $0.5

$500 × $0.5 = $250

25c = $0.25

Insurance cost on content :

($10,000 / $100) × $0.25

$100 × $0.25 = $25

Total insurance cost :

$250 + $25 = $275

Answer:

the answer would be

[tex] - 2 + \sqrt{3} [/tex]

but if you want it in decimal form it would be -0.26794919

Answer:

tan 165 =  -0.2679491924

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given,

Cost Price ( CP ) = Rs 400

Profit % = 4 %

Selling price ( SP ) = ?

now, Let's find the value of SP:

SP = [tex] \frac{CP \: (100 + \: profit percent \: )}{100} [/tex]

Plug the values

[tex] = \frac{400(100 + 4)}{100} [/tex]

Add the numbers

[tex] = \frac{400 \times 104}{100} [/tex]

Multiply the numbers

[tex] = \frac{41600}{100} [/tex]

Divide

= Rs [tex]416[/tex]

Selling price ( SP ) = Rs 416

Hope this helps...

Best regards!!

easy.This answer is 590

Answer:

{-1, 3, 7, 11, 15}

Step-by-step explanation:

we just substitute the set of even numbers in the function f(x) given to get our corresponding set of odds numbers

for 0

2(0)-1=0-1=-1

for 2

2(2)-1=4-1=3

for 4

2(4)-1=8-1=7

for 6

2(6)-1=12-1=11

for 8

2(8)-1=16-1=15