let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as exponetial distribution with mean of 4 ton per day . The amount of raw sugar (x) has

Answer:

AB = 2 sqrt(35)   (or 11.83 to two decimal places)

Step-by-step explanation:

Refer to diagram.

ABO'P is a rectangle (all angles 90)

=>

PO'  =  AB

AB = PO' = sqrt(12^2-2^2) = sqrt(144-4) = sqrt(140) = 2sqrt(35)

using Pythagoras theorem.

Answer:

(0,-2)

Step-by-step explanation:

The y-intercept is simply when the function touches or crosses the y-axis.

We're told that the graph crosses the y-axis at -2. In other words, the y-intercept is at -2.

The ordered pair would be (0,-2)

Answer:

2 inches

Step-by-step explanation:

x= smallest

3x=largest

2x=medium

x+3x+2x=12

6x=12

x=2

so smallest is 2

largest is 6 (3x)

medium is 4 (2x)

2+6+4=12

Answer:

The probability that no more than 70% would prefer to start their own business is 0.1423.

Step-by-step explanation:

We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.

Let [tex]\hat p[/tex] = sample proportion of people who prefer to start their own business

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

                               Z  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]  ~ N(0,1)

where, p = population proportion who would prefer to start their own business = 72%

            n = sample of 18-29 year-olds = 600

Now, the probability that no more than 70% would prefer to start their own business is given by = P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%)

       P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.07) = 1 - P(Z < 1.07)

                                                                       = 1 - 0.8577 = 0.1423

The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.

Answer:

8lb of the cheaper Candy

17.5lb of the expensive candy

Step-by-step explanation:

Let the cheaper candy be x

let the costly candy be y

X+y = 25.5....equation one

2.2x +7.3y = 25.5(5.7)

2.2x +7.3y = 145.35.....equation two

X+y = 25.5

2.2x +7.3y = 145.35

Solving simultaneously

X= 25.5-y

Substituting value of X into equation two

2.2(25.5-y) + 7.3y = 145.35

56.1 -2.2y +7.3y = 145.35

5.1y = 145.35-56.1

5.1y = 89.25

Y= 89.25/5.1

Y= 17.5

X= 25.5-y

X= 25.5-17.5

X= 8

Answer:

Step-by-step explanation:

The average salary in this city is $45,600.

Using the formula

z score = x - u /(sd/√n)

Where x is 46,356, u is 45,600 sd is 15,930 and n is 53.

z = 46,356 - 45600 / (15930/√53)

z = 756/(15930/7.2801)

z = 756/(2188.1568)

z = 0.3455

To draw a conclusion, we have to determine the p value, at 0.05 level of significance for a two tailed test, the p value is 0.7297. The p value is higher than the significance level, thus we will fail to reject the null and can conclude that there is not enough statistical evidence to prove that the average is any different for single people.

Answer:

6x + y = -18

Step-by-step explanation:

The given equation is,

y - 6 = -6(x + 4)

We have to rewrite this equation in the form of Ax + By = C

Where A, B and C are the integers.

By solving the given equation,

y - 6 = -6x - 24 [Distributive property]

y - 6 + 6 = -6x - 24 + 6 [By adding 6 on both the sides of the equation]

y = -6x - 18

y + 6x = -6x + 6x - 18

6x + y = -18

Here A = 6, B = 1 and C = -18.

Therefore, 6x + y = -18 will be the equation.

Answer:

In Table C, y vary inversely with x.

1×18 = 18

2×9 = 18

3×6 = 18

18 = 18 = 18

Step-by-step explanation:

We are given four tables and asked to find out in which table y vary inversely with x.

We know that an inverse relation has a form given by

y = k/x

xy = k

where k must be a constant

Table A:

x     |      y

1     |      3

2     |     9

3     |    27

1×3 = 3

2×9 = 18

3×27 = 81

3 ≠ 18 ≠ 81

Hence y does not vary inversely with x.

Table B:

x     |      y

1     |     -5

2     |     5

3     |    15

1×-5 = -5

2×5 = 10

3×15 = 45

-5 ≠ 10 ≠ 45

Hence y does not vary inversely with x.

Table C:

x     |      y

1     |      18

2     |     9

3     |     6

1×18 = 18

2×9 = 18

3×6 = 18

18 = 18 = 18

Hence y vary inversely with x.

Table D:

x     |      y

1     |      4

2     |     8

3     |    12

1×4 = 4

2×8 = 16

3×12 = 36

4 ≠ 16 ≠ 36

Hence y does not vary inversely with x.

Answer: 680

Step-by-step explanation:

When order doesn't matter,then the number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

Given: Total participants = 17

From these, a group of 3 participants is to be tested under a special condition.

Number of groups of 3 participants chosen = [tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\[/tex]

[tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\\\\=\dfrac{17\times16\times15\times14!}{3\times2\times14!}\\\\=680[/tex]

Hence, there are 680 groups of 3 participants can  be chosen,.

Answer:

Hey there!

The common ratio is 5, because you multiply by 5 to get from one term to the next.

Hope this helps :)

Answer:

5

Step-by-step explanation:

To find the common ratio take the second term and divide by the first term

55/11 = 5

The common ratio would be 5

Answer:

the probability that one parachute of the  five parachute is damaged is 0.156

Step-by-step explanation:

From the given information;

Let consider X to be the altitude above the  ground that a parachute opens

Then; we can posit that the probability that the parachute is damaged is:

P(X ≤ 100 )

Given that the population mean μ = 155

the standard deviation σ = 30

Then;

[tex]P(X \leq 100 ) = ( \dfrac{X- \mu}{\sigma} \leq \dfrac{100- \mu}{\sigma})[/tex]

[tex]P(X \leq 100 ) = ( \dfrac{X- 155}{30} \leq \dfrac{100- 155}{30})[/tex]

[tex]P(X \leq 100 ) = (Z \leq \dfrac{- 55}{30})[/tex]

[tex]P(X \leq 100 ) = (Z \leq -1.8333)[/tex]

[tex]P(X \leq 100 ) = \Phi( -1.8333)[/tex]

From standard normal tables

[tex]P(X \leq 100 ) = 0.0334[/tex]

Hence; the probability of the given parachute damaged is 0.0334

Let consider Q to be the dropped parachute

Given that the number of parachute be n= 5

The probability that the parachute opens in each trail be  p = 0.0334

Now; the random variable Q follows the binomial distribution with parameters n= 5 and p = 0.0334

The probability mass function is:

Q [tex]\sim[/tex] B(5, 0.0334)

Similarly; the event that one parachute is damaged is :

Q ≥ 1

P( Q ≥ 1 ) = 1 - P( Q < 1 )

P( Q ≥ 1 ) = 1 - P( Y = 0 )

P( Q ≥ 1 ) = 1 - b(0;5; 0.0334 )

P( Q ≥ 1 ) = [tex]1 -(^5_0)* (0.0334)^0*(1-0.0334)^5[/tex]

P( Q ≥ 1 ) = [tex]1 -( \dfrac{5!}{(5-0)!}) * (0.0334)^0*(1-0.0334)^5[/tex]

P( Q ≥ 1 ) = 1 -  0.8437891838

P( Q ≥ 1 ) = 0.1562108162

P( Q ≥ 1 ) [tex]\approx[/tex] 0.156

Therefore; the probability that one parachute of the  five parachute is damaged is 0.156

Answer:

The required probability = 0.144

Step-by-step explanation:

Since the probability of making money is 60%, then the probability of losing money will be 100-60% = 40%

Now the probability we want to calculate is the probability of making money in the first two days and losing money on the third day.

That would be;

P(making money) * P(making money) * P(losing money)

Kindly recollect;

P(making money) = 60% = 60/100 = 0.6

P(losing money) = 40% = 40/100 = 0.4

The probability we want to calculate is thus;

0.6 * 0.6 * 0.4 = 0.144

Answer:

a. p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]  

b.0.6 ±  1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]  

c. { -1.96 ≤  p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]     ≥ 1.96} = 0.95  

Step-by-step explanation:

Here the total number of trials is n= 1000

The number of successes is p` = 600/1000 = 0.6. The q` is 1 - p`= 1- 0.6 = 0.4

The degree of confidence is 95 %  therefore z₀.₀₂₅ = 1.96 ( α/2 = 0.025)

a.  The formula used will be

p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]       ( z with the base alpha by 2 (α/2 = 0.025))

b. Putting the values

0.6 ±  1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]  

c. Confidence Interval in Interval Notation.

{ -1.96 ≤  p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]     ≥ 1.96} = 0.95  

{ -z( base alpha by 2) ≤  p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]     ≥ z( base alpha by 2)  } = 1- α

Complete Question

If w'(t) is the rate of growth of a child in pounds per year, what does

[tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex]  represent?

a) The change in the child's weight (in pounds) between the ages of 4 and 7.

b) The change in the child's age (in years) between the ages of 4 and 7.

c) The child's weight at age 7.

d) The child's weight at age 4. The child's initial weight at birth.

Answer:

The correct option is  option a

Step-by-step explanation:

From the question we are told that

       [tex]w'(t)[/tex] represents the rate of growth of a child in   [tex]\frac{pounds}{year}[/tex]

So      [tex]{w'(t)} \, dt[/tex]  will be in  [tex]pounds[/tex]

Which then mean that this  [tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex]  the change in the weight of the child between the ages of  [tex]4 \to 7[/tex] years

The area of the surface above the region R is 4096π square units.

Given that:

The function: [tex]f(x, y) = 64 + x^2 - y^2[/tex]

The region R is the disk with a radius of 8 units [tex]x^2 + y^2 \le 64[/tex].

To find the area of the surface given by z = f(x, y) that lies above the region R, to calculate the double integral over the region R of the function f(x, y) with respect to dA.

The integral for the area is given by:

[tex]Area = \int\int_R f(x, y) dA[/tex]

To evaluate this integral, we need to set up the limits of integration for x and y over the region R, which is the disk cantered at the origin with a radius of 8 units.

Using polar coordinates, we can parameterize the region R as follows:

x = rcos(θ)

y = rsin(θ)

where r goes from 0 to 8, and θ goes from 0 to 2π.

Now, rewrite the integral in polar coordinates:

[tex]Area =\int\int_R f(x, y) dA\\Area = \int_0 ^{2\pi} \int_0^8(64 + r^2cos^2(\theta) - r^2sin^2(\theta)) \times r dr d \theta[/tex]

Now, we can integrate with respect to r first and then with respect to θ:

[tex]Area = \int_0^{2\pi} \int_0^8] (64r + r^3cos^2(\theta) - r^3sin^2(\theta)) dr d \theta[/tex]

Integrate with respect to r:

[tex]Area = \int_0^{2\pi}[(32r^2 + (1/4)r^4cos^2(\theta) - (1/4)r^4sin^2(\theta))]_0^8 d \theta\\Area = \int_0^{2\pi} (2048 + 256cos^2(\theta) - 256sin^2(\theta)) d \theta[/tex]

Now, we can integrate with respect to θ:

[tex]Area = [2048\theta + 128(sin(2\theta) + \theta)]_0 ^{2\pi}[/tex]

Area = 2048(2π) + 128(sin(4π) + 2π) - (2048(0) + 128(sin(0) + 0))

Area = 4096π + 128(0) - 0

Area = 4096π square units

So, the area of the surface above the region R is 4096π square units.

Learn more about Integration here:

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Answer:

33800

Step-by-step explanation:

100 x 338 = 33800

Answer:

33800

Step-by-step explanation:

338x10=3380 then 3380x10=33800

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Good luck with your assignment...

Answer:

0.0668 or 6.68%

Step-by-step explanation:

Variance (V) = 10,000

Standard deviation (σ) = √V= 100

Mean score (μ) = 500

The z-score for any test score X is:

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

For X = 650:

[tex]z=\frac{650-500}{100}\\z=1.5[/tex]

A z-score of 1.5 is equivalent to the 93.32nd percentile of a normal distribution. Therefore, the probability that he or she will make a score of 650 or more is:

[tex]P(X\geq 650)=1-P(X\leq 650)\\P(X\geq 650)=1-0.9332\\P(X\geq 650)=0.0668=6.68\%[/tex]

The probability is 0.0668 or 6.68%

The probability that he or she will make a score of 650 or more is 0.0668.

Let X = Scores made on a certain aptitude test by nursing students

X follows normal distribution with mean = 500 and variance of 10,000.

So, standard deviation = [tex]\sqrt{10000}=100[/tex].

z score of 650 is = [tex]\frac{\left(650-500\right)}{100}=1.5[/tex].

The probability that he or she will make a score of 650 or more is:

[tex]P(X\geq 650)\\=P(z\geq 1.5)\\=1-P(z<1.5)\\=1-0.9332\\=0.0668[/tex]

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Answer: The length of the line B'C" is 1 unit.

Step-by-step explanation:

Given: Triangle ABC is dilated by a scale factor of 0.5 with the origin as the center of dilation , resulting in the image Triangle A'B'C'.

If A (2,2), B= (4,3) and C=(6,3).

Distance between (a,b) and (c,d): [tex]D=\sqrt{(d-b)^2+(c-b)^2}[/tex]

Then, BC [tex]=\sqrt{(3-3)^2+(6-4)^2}[/tex]

[tex]\\\\=\sqrt{0+2^2}\\\\=\sqrt{4}\\\\=2\text{ units}[/tex]

Length of image = scale factor x length in original figure

B'C' = 0.5 × BC

= 0.5 × 2

= 1 unit

Hence, the length of the line B'C" is 1 unit.

Answer:

49.923 mph

Step-by-step explanation:

we know that the car traveled 133 miles in h hours at an average speed of x mph.

That is, xh = 133.

We can also write this in terms of hours driven: h = 133/x.

If x was 30 mph faster, then h would be one hour less.

That is, (x + 30)(h - 1) = 133, or h - 1 = 133/(x + 30).

We can rewrite the latter equation as h = 133/(x + 30) + 1

We can then make a system of equations using the formulas in terms of h to find x:

h = 133/x = 133/(x + 30) + 1

133/x = 133/(x + 30) + (x + 30)/(x + 30)

133/x = (133 + x + 30)/(x + 30)

133 = x*(133 + x + 30)/(x + 30)

133*(x + 30) =  x*(133 + x + 30)

133x + 3990 = 133x + x^2 + 30x

3990 = x^2 + 30x

x^2 + 30x - 3990 = 0

Using the quadratic formula:

x = [-b ± √(b^2 - 4ac)]/2a  

= [-30 ± √(30^2 - 4*1*(-3990))]/2(1)  

= [-30 ± √(900 + 15,960)]/2

= [-30 ± √(16,860)]/2

= [-30 ± 129.846]/2

= 99.846/2  -----------  x is miles per hour, and a negative value of x is neglected, so we'll use the positive value only)

= 49.923

Check if the answer is correct:

h = 133/49.923 = 2.664, so the car took 2.664 hours to drive 133 miles at an average speed of 49.923 mph.

If the car went 30 mph faster on average, then h = 133/(49.923 + 30) = 133/79.923 = 1.664, and 2.664 - 1 = 1.664.

Thus, we have confirmed that a car driving 133 miles at about 49.923 mph would have arrive precisely one hour earlier by going 30 mph faster

Answer:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = [tex]\frac{1}{4}[/tex]

Step-by-step explanation:

Given that

3 white, 2 blue and 5 gray shirts are there.

To find:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?

Solution:

Here, total number of shirts = 3+2+5 = 10

First of all, let us learn about the formula of an event E:

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]

[tex]P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}[/tex]

[tex]P(First\ White) = \dfrac{3}{10}[/tex]

Now, this shirt is set aside.

So, total number of shirts left are 9 now.

[tex]P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }[/tex]

So, the answer is:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = [tex]\frac{1}{4}[/tex]

Answer:

a) $3700

b) $555

Step-by-step explanation:

The length of the loan is 3 years.

The interest after 3 years is $444.

The rate of the Simple Interest is 4%.

Simple Interest is given as:

I = (P * R * T) / 100

where P = principal (amount borrowed)

R = rate

T = length of years

Therefore:

[tex]444 = (P * 3 * 4) / 100\\\\444 = 12P / 100\\\\12P = 444 * 100\\\\12P = 44400\\\\P = 44400 / 12\\[/tex]

P = $3700

She borrowed $3700

b) If the simple interest was 5%, then:

I = (3700 * 5 * 3) / 100 = $555

The interest would be $555.

I hope this will help uh.....

Answer:

(a) The probability of getting someone who was not sent to prison is 0.55.

(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is 0.63.

Step-by-step explanation:

We are given that in a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.

Let the probability that subjects studied were sent to prison = P(A) = 0.45

Let G = event that subject chose to plead guilty

So, the probability that the subjects chose to plead guilty given that they were sent to prison = P(G/A) = 0.40

and the probability that the subjects chose to plead guilty given that they were not sent to prison = P(G/A') = 0.55

(a) The probability of getting someone who was not sent to prison = 1 - Probability of getting someone who was sent to prison

      P(A') = 1 - P(A)

               = 1 - 0.45 = 0.55

(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is given by = P(A'/G)

We will use Bayes' Theorem here to calculate the above probability;

    P(A'/G) =  [tex]\frac{P(A') \times P(G/A')}{P(A') \times P(G/A') +P(A) \times P(G/A)}[/tex]      

                 =  [tex]\frac{0.55 \times 0.55}{0.55\times 0.55 +0.45 \times 0.40}[/tex]

                 =  [tex]\frac{0.3025}{0.4825}[/tex]

                 =  0.63

You want to end up with [tex]A\sin(\omega t+\phi)[/tex]. Expand this using the angle sum identity for sine:

[tex]A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi[/tex]

We want this to line up with [tex]2\sin(4\pi t)+5\cos(4\pi t)[/tex]. Right away, we know [tex]\omega=4\pi[/tex].

We also need to have

[tex]\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}[/tex]

Recall that [tex]\sin^2x+\cos^2x=1[/tex] for all [tex]x[/tex]; this means

[tex](A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}[/tex]

Then

[tex]\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)[/tex]

So we end up with

[tex]2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)[/tex]

Answer:

Step-by-step explanation:

In the form ...

  y(t) = Asin(ωt +φ)

you have ...

The translation from ...

  y(t) = 2sin(4πt) +5cos(4πt)

is ...

  A = √(2² +5²) = √29 . . . . the amplitude

  ω = 4π . . . . the angular frequency in radians per second

  φ = arctan(5/2) ≈ 1.1903 . . . . radians phase shift

Then, ...

  y(t) = √29·sin(4πt +1.1903)

_____

Comment on the conversion

You will notice we used "2" and "5" to find the amplitude and phase shift. In the generic case, these are "coefficient of sin( )" and "coefficient of cos( )". When determining phase shift, pay attention to whether your calculator is giving you degrees or radians. (Set the mode to what you want.)

If you have a negative coefficient for sin( ), you will need to add 180° (π radians) to the phase shift value given by the arctan( ) function.

Answer:

Step-by-step explanation:

The class interval is simply the difference between the lower or upper class boundary or limit  of a class and the lower or upper class boundary or limit of the next class.

In this case for the class

$4 up to $7 18 and

$7 up to $10 36

The lower class boundary of the first class is $4 and the lower class boundary of the second class is $7

Answer: Price-earnings ratio= 22.0

Step-by-step explanation:

Given: A company had a market price of $38.50 per share, earnings per share of $1.75, and dividends per share of $0.90

To find: price-earnings ratio

Required formula: [tex]\text{price-earnings ratio }=\dfrac{\text{ Market Price per Share}}{\text{Earnings Per Share}}[/tex]

Then, Price-earnings ratio = [tex]\dfrac{\$38.50}{\$1.75}[/tex]

⇒Price-earnings ratio = [tex]\dfrac{22}{1}[/tex]

Hence, the price-earnings ratio= 22.0

Answer:

z(c)  = - 1,64

We reject the null hypothesis

Step-by-step explanation:

We need to solve a proportion test ( one tail-test ) left test

Normal distribution

p₀ = 63 %

proportion size  p = 51 %

sample size  n = 114

At 5% level of significance   α = 0,05, and with this value we find in z- table z score of z(c) = 1,64  ( critical value )

Test of proportion:

H₀     Null Hypothesis                        p = p₀

Hₐ    Alternate Hypothesis                p < p₀

We now compute z(s) as:

z(s) =  ( p - p₀ ) / √ p₀q₀/n

z(s) =( 0,51 - 0,63) / √0,63*0,37/114

z(s) =  - 0,12 / 0,045

z(s) = - 2,66

We compare z(s) and z(c)

z(s) < z(c)      - 2,66 < -1,64

Therefore as z(s) < z(c)  z(s) is in the rejection zone we reject the null hypothesis

Answer:

It is unlikely.

Step-by-step explanation:

Certain = 100%

Impossible = 0%

Likely = more than 50%

Unlikely = less than 50%

It is less than 50%, so it is unlikely.

Answer:

(A) it is likely

Step-by-step explanation:

i took the test on edge

Answer:

j < 2

Step-by-step explanation:

Simplify both sides of the inequality and isolating the variable would get you the answer

Answer:

[tex]\boxed{x = 9}[/tex]

Step-by-step explanation:

m = -1/3

b = 7

And y = 4 (Given)

Putting all of the givens in [tex]y = mx+b[/tex] to solve for x

=> 4 = (-1/3) x + 7

Subtracting 7 to both sides

=> 4-7 = (-1/3) x

=> -3 = (-1/3) x

Multiplying both sides by -3

=> -3 * -3 = x

=> 9 = x

OR

=> x = 9

Answer:

x = 9

Step-by-step explanation:

m = -1/3

b = 7

Using slope-intercept form:

y = mx + b

m is slope, b is y-intercept.

y = -1/3x + 7

Solve for x:

Plug y as 4

4 = 1/3x + 7

Subtract 7 on both sides.

-3 = -1/3x

Multiply both sides by -3.

9 = x