Which value of y makes the equation y/9=12 true

Answer:

[tex]t \approx 17.690\,min[/tex]

Step-by-step explanation:

This differential equation is a first order linear differential equation with separable variables, whose solution is found as follows:

[tex]\frac{dy}{dt} = - \frac{1}{53} \cdot (y - 17)[/tex]

[tex]\frac{dy}{y-17} = -\frac{1}{53} \, dt[/tex]

[tex]\int\limits^{y}_{y_{o}} {\frac{dy}{y-17} } = -\frac{1}{53} \int\limits^{t}_{0}\, dx[/tex]

[tex]\ln \left |\frac{y-17}{y_{o}-17} \right | = -\frac{1}{53} \cdot t[/tex]

[tex]\frac{y-17}{y_{o}-17} = e^{-\frac{1}{53}\cdot t }[/tex]

[tex]y = 17 + (y_{o} - 17) \cdot e^{-\frac{1}{53}\cdot t }[/tex]

The solution of the differential equation is:

[tex]y = 17 + 74\cdot e^{-\frac{1}{53}\cdot t }[/tex]

Where:

[tex]y[/tex] - Temperature, measured in °C.

[tex]t[/tex] - Time, measured in minutes.

The time when the cup of coffee has the temperature of 70 °C is:

[tex]70 = 17 + 74 \cdot e^{-\frac{1}{53}\cdot t }[/tex]

[tex]53 = 74 \cdot e^{-\frac{1}{53}\cdot t }[/tex]

[tex]\frac{53}{74} = e^{-\frac{1}{53}\cdot t }[/tex]

[tex]\ln \frac{53}{74} = -\frac{1}{53}\cdot t[/tex]

[tex]t = - 53\cdot \ln \frac{53}{74}[/tex]

[tex]t \approx 17.690\,min[/tex]

Answer:

The approximate probability that more than six students were born on Christmas day is P=0.105.

Step-by-step explanation:

This can be modeled as a binomial variable, with n=1460 and p=1/365.

The sample size n is the total amount of students and the probability of success p is the probability of each individual of being born on Christmas day.

As the sample size is too large to compute it as a binomial random variable, we approximate it to the normal distribution with the following parameters:

[tex]\mu=n\cdot p=1460\cdot (1/365)=4\\\\\sigma=\sqrt{n\cdot p(1-p)}=\sqrt{1460\cdot(1/365)\cdot(364/365)}=\sqrt{3.989}=1.997[/tex]

We want to calculate the probability that more than 6 students were born on Christmas day. Ww apply the continuity factor and we write the probability as:

[tex]P(X>6.5)[/tex]

We calculate the z-score for X=6.5 and then calculate the probability:

[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{6.5-4}{1.997}=\dfrac{2.5}{1.997}=1.252\\\\\\P(X>6.5)=P(z>1.252)=0.105[/tex]

Answer: Choice C

Step-by-step explanation:

Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true.

3/10≠3/5*1/4

so event A and B are not independent.

[tex]answer \\ 13.5 \: m/s \\ given \\ initial \: velocity(u) = \: 3 \: m/s \\ acceleration(a) = 1.5 \: m/s ^2\\ time(t) = 7 \: second \\ now \\ v = u + at \\ or \: v = 3 + 1.5 \times 7 \\ \: or \: v = 3 + 10.5 \\ v = 13.5 \: m/s \\ hope \: it \: helps[/tex]

Answer:

[tex]y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ][/tex]

Step-by-step explanation:

Solution:-

- Given is the 2nd order linear ODE as follows:

                      [tex]ty'' + ( 2t - 1 )*y' - 2y = 6t^2 . e^(^-^2^t^)[/tex]

- The complementary two independent solution to the homogeneous 2nd order linear ODE are given as follows:

                     [tex]y_1(t) = 2t - 1\\\\y_2 (t ) = e^-^2^t[/tex]

- The particular solution ( yp ) to the non-homogeneous 2nd order linear ODE is expressed as:

                    [tex]y_p(t) = u_1(t)*y_1(t) + u_2(t)*y_2(t)[/tex]

Where,

              [tex]u_1(t) , u_2(t)[/tex] are linearly independent functions of parameter ( t )

- To determine [  [tex]u_1(t) , u_2(t)[/tex] ], we will employ the use of wronskian ( W ).

- The functions [[tex]u_1(t) , u_2(t)[/tex] ] are defined as:

                       [tex]u_1(t) = - \int {\frac{F(t). y_2(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\\\u_2(t) = \int {\frac{F(t). y_1(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\[/tex]

Where,

      F(t): Non-homogeneous part of the ODE

      W [ y1(t) , y2(t) ]: the wronskian of independent complementary solutions

- To compute the wronskian W [ y1(t) , y2(t) ] we will follow the procedure to find the determinant of the matrix below:

                      [tex]W [ y_1 ( t ) , y_2(t) ] = | \left[\begin{array}{cc}y_1(t)&y_2(t)\\y'_1(t)&y'_2(t)\end{array}\right] |[/tex]

                      [tex]W [ (2t-1) , (e^-^2^t) ] = | \left[\begin{array}{cc}2t - 1&e^-^2^t\\2&-2e^-^2^t\end{array}\right] |\\\\W [ (2t-1) , (e^-^2^t) ]= [ (2t - 1 ) * (-2e^-^2^t) - ( e^-^2^t ) * (2 ) ]\\\\W [ (2t-1) , (e^-^2^t) ] = [ -4t*e^-^2^t ]\\[/tex]

- Now we will evaluate function. Using the relation given for u1(t) we have:

                     [tex]u_1 (t ) = - \int {\frac{6t^2*e^(^-^2^t^) . ( e^-^2^t)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_1 (t ) = \frac{3}{2} \int [ t*e^(^-^2^t^) ] \, dt\\\\u_1 (t ) = \frac{3}{2}* [ ( -\frac{1}{2} t*e^(^-^2^t^) - \int {( -\frac{1}{2}*e^(^-^2^t^) )} \, dt] \\\\u_1 (t ) = -e^(^-^2^t^)* [ ( \frac{3}{4} t + \frac{3}{8} )] \\\\[/tex]

- Similarly for the function u2(t):

                     [tex]u_2 (t ) = \int {\frac{6t^2*e^(^-^2^t^) . ( 2t-1)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_2 (t ) = -\frac{3}{2} \int [2t^2 -t ] \, dt\\\\u_2 (t ) = -\frac{3}{2}* [\frac{2}{3}t^3 - \frac{1}{2}t^2 ] \\\\u_2 (t ) = t^2 [\frac{3}{4} - t ][/tex]

- We can now express the particular solution ( yp ) in the form expressed initially:

                    [tex]y_p(t) = -e^(^-^2^t^)* [\frac{3}{2}t^2 + \frac{3}{4}t - \frac{3}{8} ] + e^(^-^2^t^)*[\frac{3}{4}t^2 - t^3 ]\\\\y_p(t) = -e^(^-^2^t^)* [t^3 + \frac{3}{4}t^2 + \frac{3}{4}t - \frac{3}{8} ] \\[/tex]

Where the term: 3/8 e^(-2t) is common to both complementary and particular solution; hence, dependent term is excluded from general solution.

- The general solution is the superposition of complementary and particular solution as follows:

                    [tex]y_g(t) = y_c(t) + y_p(t)\\\\y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ][/tex]

Answer:

⬇⬇⬇⬇⬇⬇

⬇⬇⬇⬇⬇⬇

Step-by-step explanation:

1, 3, 4

proof below

(1) The area of the rectangle is 88 square inches

(3) The equation x² – 3x – 88 = 0 can be used to solve for the length of the rectangle.

(4) The triangle has a base of 11 inches and a height of 8 inches.

Area of a rectangle is the sum of the area of two equal right triangle.

Area of rectangle = 2(area of right triangle)

Area of rectangle = 2(44 sq inches) = 88 sq inches

Let the length = x

then the width becomes, x - 3

Area = x(x - 3) = 88

x² - 3x = 88

x² - 3x - 88 = 0

x = 11

width = 11 - 3 = 8

Thus, the statements that are true include;

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Answer:

The answer is No Solution

Answer:

No solution

Step-by-step explanation:

There is no solution to this question.

Since the x is an absolute number, the answer cannot be -4. It would have to equal 4

So if that x was a -4 it would equal 4 since it is in absolute value brackets

Answer:

they played the pivotal role

Answer:

Option (2).

Step-by-step explanation:

From the figure attached,

EFGH is a quadrilateral and FH is line which divides the quadrilateral into two right triangles, ΔFEH and ΔHGF.

In ΔFEH and ΔHGF,

Sides EH ≅ FG [Given]

FH ≅ FH [reflexive property]

ΔFEH ≅ ΔHGF [HL (Hypotenuse - length) postulate of congruence]

Option (2) will be the answer.

Answer:

D. $7.50

Step-by-step explanation:

$157.50 / 21 hours = $7.50 per hour

Answer:

30 years old

Step-by-step explanation:

Joe's age can be set to x. Since Billy is twice as old as Joe, his age can be set to 2x. They add up to 45. You can set up an equation, 2x+x=45.

Simplify the equation.

3x=45, x=15

Joe is 15. Billy is 2(15)=30.

Answer:

n = -1

Step-by-step explanation:

So first subtract 9 to both sides

8n = -n - 9

Now you want the n on one side and the constant on the other

so add the single n to the n side

9n = -9

Divide 9 to both sides to solve for n

n = -1

Answer:

  C

Step-by-step explanation:

The function of table A can be written as ...

  y = 100x -92

__

The function of table B can be written as ...

  y = 10x +446

__

The function of table C can be written as ...

  y = (5/3)·3^x

__

The function of table D can be written as ...

  y = 2x +413

__

The exponential function of Table C will have the largest y-values for any value of x greater than 6.

_____

Comment on the functions

When trying to determine the nature of the function, it is often useful to look at the differences of the y-values for consecutive x-values. Here, the first-differences are constant for all tables except C. That means functions A, B, D are linear functions.

If the first differences are not constant, one can look at second differences and at ratios. For table C, we notice that each y-value is 3 times the previous one. That constant ratio means the function is exponential, hence will grow faster than any linear function.

Answer:

yes, what the other user  is correct i just took the quiz

Step-by-step explanation:

Answer:

31 degrees

Step-by-step explanation:

Since RPS and QPR make up QPS, the sum of their angle measures must be 47. Therefore:

3x-38+7x-95=47

10x-133=47

10x=180

x=18

QPR=7(18)-95=126-95=31

Hope this helps!

Answer:

4.700

Step-by-step explanation:

Find the number in the thousandth place  0  and look one place to the right for the rounding digit 4. Round up if this number is greater than or equal to 5 and round down if it is less than 5.

brainliesssss plssssssssssssss

Answer:

Radius = 13 m

Step-by-step explanation:

Formula for area of circle is given as:

[tex]A = \pi {r}^{2} \\ \\ \therefore \: 169\pi \: = \pi {r}^{2} \\ \\ \therefore \: {r}^{2} = \frac{169\pi }{\pi} \\ \\ \therefore \: {r}^{2} = 169 \\ \\ \therefore \: {r} = \pm \sqrt{169} \\ \\\therefore \: r = \pm \: 13 \: m \\ \\ \because \: radius \: of \: a \: circle \: can \: not \: be \: a \: negative \: \\quantity \\ \\ \huge \red{ \boxed{\therefore \: r = 13 \: m }}[/tex]

Answer: -1

Step-by-step explanation:

You want to find an angle that is coterminal to 495. So, subtract 360 degrees until youre in the range of 0-360. I got 495 - 360 = 135°

Tangent is equal to [tex]\frac{sin(theta)}{cos(theta)}[/tex], we already solved theta which was 135°

This next part is hard to explain to someone who doesnt know their trig circle, idk if you do. The angle 135 is apart of the pi/4 gang. So we know this is going to be some variant of √2/2. Sine of quadrant 1 and 2 is gonna be positive:

[tex]sin135=\frac{\sqrt{2} }{2}[/tex]

Now lets do cosine of 135°, which again is apart of the pi/4 gang because its divisible by 45°. Its in quadrant 2 so the cosine will be negative.

[tex]cos135=-\frac{\sqrt{2} }{2}[/tex]

The final step is to divide them. They are both fractions so you should multiply by the reciprocal.

[tex]\frac{\sqrt{2} }{2} *-\frac{2}{\sqrt{2} } =-\frac{2\sqrt{2} }{2\sqrt{2} } =-1[/tex]

Answer:

Yes, it would be unusual.

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 normally distributed samples are 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.

If [tex]Z \leq -2[/tex] or [tex]Z \geq 2[/tex], the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this question, we have that:

[tex]\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1[/tex]

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

[tex]Z = \frac{19 - 22}{1}[/tex]

[tex]Z = -3[/tex]

[tex]Z = -3 \leq -2[/tex], so yes, the sample mean being less than 19 days would be considered an unusual outcome.

Answer:

D

Step-by-step explanation:

Length × Width = Area

So we'll substitute the Area of the circle having formula, πr²

Answer:

The one with the better deal would be 30 ride tickets for $22.50 this is because you pay less money for more rides.

Step-by-step explanation:

First you divide 20 by 14. Doing this will give you the cost of a ride per ticket.

20/14 = 1.42

Then you do the same thing to 30 and 22.50.

30/22.50 = 1.30

Last you compare which deal has less money per ride.

1.42 > 1.30

Answer:

Step-by-step explanation:

Let x be the random variable representing the SAT scores for the students at a local high school. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 1527

σ = 291

the probability to be determined is expressed as P(x > 1207)

P(x > 1207) = 1 - P(x ≤ 1207)

For x < 1208

z = (1207 - 1527)/291 = - 1.1

Looking at the normal distribution table, the probability corresponding to the z score is 0.16

P(x > 1207) = 1 - 0.16 = 0.84

Therefore, the percentage of students from this school earn scores that satisfy the admission requirement is

0.84 × 100 = 84%

Answer:

Step-by-step explanation:

To find the average rate of change, find the change in function value, and divide that by the length of the interval.

1. ((g(1) -g(-1))/(1 -(-1)) = ((-4·1³ +4) -(-4(-1)³ +4)/(2) = (-8)/2 = -4

The average rate of change of g(x) on [-1, 1] is -4.

__

2. ((g(3) -g(-2))/(3 -(-2)) = ((6·3³ +3/3²) -(6·(-2)³ +3/(-2)²))/5

  = (6·27 +1/3 -6·(-8) -3/4)/5 = (2515/12)/5

  = 503/12 = 41 11/12

The average rate of change of g(x) on [-2, 3] is 41 11/12.

Answer:

y= -3/2x+12

Step-by-step explanation:

the slope of perpendicular lines multiplied together would be -1, so the slope of the perpendicular line is -3/2. y=-3/2x+b, so -6=-18+b, so b= 12. the equation of the line is y=-3/2x+12.

Answer:

0.347% of the total tires will be rejected as underweight.

Step-by-step explanation:

For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.

And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.

1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344

1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792

The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)

Using data from the normal distribution table

P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight

Hope this Helps!!!

The proportion of the tires that would be denied for being underweight through the given process would be:

[tex]0.347[/tex]% of the total tires will be rejected as underweight.

Given that,

Interquartile Range [tex]= 1.5[/tex]

Standard Deviation [tex]= 0.76[/tex]

Considering Mean = 0

and Standard Deviation = 1

Since lower quartile = -0.67448

Upper quartile  = +0.67448

IQ range =  1.34896

To find,

The proportion of tires would be rejected due to being underweight through the process would be:  

1.5 of Interquartile Range = 1.5 × [tex]1.34896 = 2.02344[/tex]

Now,

1.5 of the IQ range below the lower quartile [tex]= (lower quartile) - (1.5 of Interquartile range)[/tex]

[tex]= -0.67448 - 2.02344[/tex]

[tex]= -2.69792[/tex]

The proportion of tires that would be under 1.5 of the interquartile range below the lower quartile:

[tex]= P(x < -2.69792)[/tex] ≈ [tex]P(x < -2.70)[/tex]

Using data through the Normal Distribution Table,

[tex]P(x < -2.70)[/tex] [tex]= 0.00347[/tex]

[tex]= 0.347[/tex]%

Thus, 0.347% of the total tires would be rejected as underweight.

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Answer:2x−5=−5

Add 5

to both sides of the equation.

2x=−5+5

Add −5

and 5

.

2x=0

Divide each term by 2

and simplify.

Divide each term in 2x=0

by 2

.

2x2=02

Cancel the common factor of 2

.

Cancel the common factor.

2

x2=02

Step-by-step explanation:

Apply the distributive property.

x⋅2+2⋅2−9=−5

Move 2

to the left of x

.

2⋅x+2⋅2−9=−5

Multiply 2

by 2

.

2x+4−9=−5

Subtract 9

from 4

.

Answer:

You are 6

Step-by-step explanation:

8×6=48 and 6/3=2

6 is even and fits in all of these areas.

Hope this helps.

Mark brainliest if correct.

Answer:

207.24cm

Step-by-step explanation:

Circumference=2pi*r

=2 (3.14)(33)

=207.24cm

The circumference of the circle is 103.62 cm².

Given that, a circle with diameter of 33 cm.

Radius=d/2=16.5 cm.

The formula to find the circumference of the circle is 2πr.

Now, 2×3.14×16.5=103.62 cm².

Therefore, the circumference of the circle is 103.62 cm².

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Answer:

2

Step-by-step explanation:

Set the height of the bar to 2 since there are 2 numbers between 21-40.

Answer:

2 people.

Step-by-step explanation:

34 minutes and 40 minutes were recorded.

Therefore, 2 people.

Answer:

y = -z/2 - x/2

Step-by-step explanation:

2y + 5x – z = 4y + 6x

-5x. -5x

2y - z = 4y + x

+z. +z

2y = 4y + z + x

-4y -4y

-2y = z + x

÷-2. ÷-2

y = -z/2 - x/2