A line with a slope of 4 passes through the point (2, 3) . What is its equation in slope -intercept form?

Answer:

(3x+2) by (2x+1)

Step-by-step explanation:

A cardboard is a rectangle, and has two dimensions. Given a quadratic equation, you should find a way to split it in two.

The easiest way to do so is through factoring. (There are many ways to do this, take a look at the plethora of sources offered on the internet.)

When the expression 6x^2 + 7x + 2 is factored, it is (3x+2)(2x+1). Hence, these are your dimensions.

Answer:

Step-by-step explanation:

This is nice and simple. I'm going to walk through it like I do when teaching this concept to my class for the first time. This is a good problem for that.

We are given a square and we are looking for the rate at which the area is increasing when a certain set of specifics are given. That means that the main equation for this problem is the area of a square, which is:

[tex]A=s^2[/tex] where s is a side.

Since we are looking for the rate at which the area is changing, [tex]\frac{dA}{dt}[/tex], we need to take the derivative of area formula implicitly:

[tex]\frac{dA}{dt}=2s\frac{ds}{dt}[/tex] that means that if [tex]\frac{dA}{dt}[/tex] is our unknown, we need values for everything else. We are given that the initial area for the square is 49. That will help us determine what the "s" in our derivative is. We plug in 49 for A and solve:

[tex]49=s^2[/tex] so

s = 7

We are also given at the start that the sides of this square are increasing at a rate of 8cm/s. That is [tex]\frac{ds}{dt}[/tex]. Filling it all in:

[tex]\frac{dA}{dt}=2(7)(8)[/tex] and

[tex]\frac{dA}{dt}=112\frac{cm^2}{s}[/tex]

The surface area of a square of side L is given by

[tex]A = L^2[/tex]

The rate of change of the area per unit time is

[tex]\dfrac{dA}{dt} = 2L\dfrac{dL}{dt}[/tex]

We can express the length L on the right hand side in terms of the area A [tex](L = \sqrt{A})[/tex]:

[tex]\dfrac{dA}{dt} = 2\sqrt{A}\dfrac{dL}{dt}[/tex]

[tex]\:\:\:\:\:\:\:=2(\sqrt{49\:\text{cm}^2})(8\:\text{cm/s})[/tex]

[tex]\:\:\:\:\:\:\:=112\:\text{cm}^2\text{/s}[/tex]

Answer:

Symmetric with respect to the x-axis

Symmetric with respect to the y-axis

Symmetric with respect to the origin

Let f(x) = ax³ + 9x² + 4x – 10

g(x) = 0⇒x - 3 = 0

⇒ x = 0 + 3

⇒ x = 3

On dividing f(x) by x - 3, it leaves a remainder 5.

Now keeping, f(3) = 5

⇒a(3)³ + 9(3)² + 4(3) - 10 = 5

⇒ a × 27 + 9 × 9 + 4 × 3 - 10 = 5

⇒ 27a + 81 + 12 - 10 = 5

⇒ 81 + 12 - 10 - 5 = 27a

⇒ 81 + 12 - 15 = 27a

⇒ 93 - 15 = 27a

⇒ 78 = 27a

⇒ a = 27/78

⇒ a = 0.3461

Answer:

See Below.

Step-by-step explanation:

We are given that:

[tex]\displaystyle I = I_0 e^{-kt}[/tex]

Where I₀ and k are constants.

And we want to prove that:

[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]

From the original equation, take the derivative of both sides with respect to t. Hence:

[tex]\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right][/tex]

Differentiate. Since I₀ is a constant:

[tex]\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)[/tex]

Using the chain rule:

[tex]\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right) = -kI_0e^{-kt}[/tex]

We have:

[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]

Substitute:

[tex]\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0[/tex]

Distribute and simplify:

[tex]\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0[/tex]

Hence proven.

Answer:

See answers below

Step-by-step explanation:

Given the following functions:

r(x) = x - 6

s(x) = 2x²

r(s(x)) = r(2x²)

Replacing x with 2x² in r(x) will give;

r(2x²) = 2x² - 6

r(s(x)) = 2x² - 6

(r-s)(x) = r(x) - s(x)

(r-s)(x) = x - 6 - 2x²

Rearrange

(r-s)(x) = - 2x²+x-6

(r+s)(x) = r(x) + s(x)

(r-s)(x) = x - 6 + 2x²

Rearrange

(r-s)(x) = 2x²+x-6

Answer:

Step-by-step explanation :

Let % increase in administrative secretary be = x

Let % increase in new faculty receive be = y

Administrative secretary salary = 28,000

New faculty receive Salary = 40,000

(8)*(x/100)* (28000) + (7)*(y/100)*(40000) = 5,000

2240x +2800 y = 5,000

224x +280 y = 500

56x +70y = 125

Therefore, x and y should be chosen such that it satisfy the above equation.

Answer:

t = 2

Step-by-step explanation:

5(6t-3)=5t+35

Step 1 distribute the 5 by multiplying 5 to what is inside of the parenthesis

5(6t) - 5(3) = 5t + 35

Outcome: 30t - 15 = 5t + 35

Step 2 add 15 to both sides

30t - 15 + 15 = 5t + 35 + 15

Outcome: 30t = 5t + 50

Step 3 subtract 5t from both sides

30t - 5t = 5t - 5t + 50

Outcome: 25t = 50

Step 4 divide both sides by 25

25t/25 = 50 we're left with t = 2

Answer:

[tex]7x {}^{2} + 5x[/tex]

Step-by-step explanation:

[tex]9x {}^{2} + 8x - (2x {}^{2} + 3x) \\ \\ = 9x {}^{2} + 8x - 2x {}^{2} - 3x (remove \: brackets) \\\ \\ = 7x {}^{2} - 5x [/tex]

The question is incomplete. The complete question is :

In a​ poll, 1100 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 43​% of respondents never​ clothes-shop online. Find and interpret a ​95% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.

Solution :

95% confidence interval for p is :

[tex]$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$[/tex]

[tex]$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$[/tex]

0.401 < p < 0.459

Therefore, 95% confidence interval is from 0.401 to 0.459

Answer:

i think its the 3 line. they are congruent.

Answer:

its the 3rd option

Step-by-step explanation:

first of all AA means angle-angle which means we are using their angles to compare them

second, those lines on the sides are only there to tell you they are in the exact same angle, and the two boxes on the bottom show that the angle of both triangles is 90° therefore they are the same

Answer:

h = 6.2 units

Step-by-step explanation:

Given triangle ABC is a right triangle with the measures of the two sides,

BC = [tex]\frac{10}{2}[/tex] = 5 units

AC = 8 units

By applying Pythagoras theorem in the given triangle,

AC² = AB² + BC²

8² = AB² + 5²

AB² = 64 - 25

AB = √39

AB = 6.24 units

AB ≈ 6.2 units

Answer:

5 cases

10 additional units

Step-by-step explanation:

Given that :

Total number of units needed = 60 units

Total number of units per case = 14

Hence, the total number of cases required will be :

Number of units needed / number of units per case

Number of cases required = 60 / 14 = 4.285 (this means that 5 cases are required as 4 cases won't be up to 60 units)

With 5 cases, we have exceeded the required units needed :

Additional units will be : (14 * 5) - 60

Additional units = 70 - 60 = 10 units

Answer:

didn't get the question did u forget to put the sequence ???????

pls write the question fully so that I can help you

Answer:

I think the answer is 5/6

Step-by-step explanation:

There are three even numbers and two uneven numbers less than four. Therefore, on a standard die, the.probability of Rollin a number that is even or less than for is 5/6.

9514 1404 393

Answer:

Step-by-step explanation:

To find the initial amount, put 0 where t is in the formula and do the arithmetic.

  A(0) = 523(1/2)^0 = 523(1) = 523

The initial amount is 523 grams.

__

To find the amount remaining after 100 years, put 100 where t is in the formula and do the arithmetic.

  A(100) = 523(1/2)^(100/30) ≈ 523(0.0992123) ≈ 52

About 52 grams will remain after 100 years.

Step-by-step explanation:

1. The inputs are the dice values. Since the cube is six sided, the possible. values are (1,2,3,4,5,6).

2. The outputs values are points if we roll the number. If we land in a special space, we must respect that rule so our outputs are

(2,2,8,5,6,8).

3. I cant show a mapping diagram on brainly. Draw a mapping diagram and make sure to connect the x values and y values of the following.Also make sure to start on square 1.

4. The domain of the function is the same as the input. The dice values, 1,2,3,4,5,6

5.The range are the. values that occur if we roll the number about square 1.

2,2,8,5,6,8

6. No, the player could have rolled 3 and landed on 8. The player also could have rolled 6.

7. Yes, every one x value corresponds with one y value.

Answer:

u = 2

Step-by-step explanation:

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

cos theta = adj side / hypotenuse

cos 45 = sqrt(2) / u

u cos 45 = sqrt(2)

u = sqrt(2) / cos 45

u = sqrt(2) / 1/ sqrt(2)

u = sqrt(2) * sqrt(2)

u =2

u=2

Answer:

Solution given:

Cos 45°=base/hypotenuse

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

doing crisscrossed multiplication

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

u=2

Explanation:

P(A) represents the probability of event A

P(not A) is the probability that event A doesn't happen

We only have two choices: Either A happens or it doesn't

So that means P(A) + P(not A) = 1

The "1" represents a 100% chance, aka certainty.

An

Step-by-step explon:

From the test the person wants, and the sample data, we build the test hypothesis, find the test statistic,  and use this to reach a conclusion both using the critical value and the p-value.

Doing this, the conclusions are:

---------------------------------------------------------

Hypothesis:

A machine in a factory must be repaired if it produces more than 10% defectives in production.

At the null hypothesis, we test if it does not have to be repaired, that is,  the proportion is of at most 10%. So

[tex]H_0: p \leq 0.1[/tex]

At the alternative hypothesis, we test if it does have to be repaired, that is, the proportion is greater than 10%. So

[tex]H_1: p > 0.1[/tex]

------------------------------------------------------

Decision rule:

0.01 significance level, using a left-tailed test(testing if the mean is more than a value), which means that:

----------------------------------------------------------

The test statistic is:

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

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.1 is tested at the null hypothesis:

This means that [tex]\mu = 0.1, \sigma = \sqrt{0.1*0.9}[/tex]

A random sample of 100 items from a day's production contains 15 defectives.

This means that [tex]n = 100, X = \frac{15}{100} = 0.15[/tex]

Value of the test-statistic:

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

[tex]z = \frac{0.15 - 0.1}{\frac{\sqrt{0.1*0.9}}{\sqrt{100}}}[/tex]

[tex]z = 1.67[/tex]

----------------------------------------------

Decision: Critical region

The test statistic is [tex]z = 1.67 < z_c[/tex], meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

Decision: p-value

The p-value of the test is the probability of finding a sample proportion above 0.15, which is 1 subtracted by the p-value of z = 1.67.

Looking at the z-table, z = 1.67 has a p-value of 0.9525.

1 - 0.9525 = 0.0475.

The p-value of the test is 0.0475 > 0.01, meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

A similar problem is found at https://brainly.com/question/24326664

Answer:

The total number of ways is 252.

Step-by-step explanation:

Number of girls = 18

number of boys = 14

Commitee of one girl and a boy

(18 C 1)(14 C 1)

= 252

Answer:

0.000000073

Step-by-step explanation:

Given number is,

7.3E - 8

In scientific notation number will be,

7.3 × 10⁻⁸

In standard form the number will be,

0.000000073

9514 1404 393

Explanation:

There are several ways you can go at this. Here are a couple. All proofs will start with the given relations being repeated as part of the proof. Here are the next steps.

∠AED ≅ ∠BAE . . . . alternate interior angles are congruent

∠AED +∠AEB = 90° . . . . angle sum theorem

∠BAE +∠AEB = 90° . . . . substitution property of equality

∠CEF ≅ ∠BCE . . . . alternate interior angles are congruent

∠CEF +∠CEB = 90° . . . . angle sum theorem

∠BCE +∠CEB = 90° . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠CEB . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠AEB . . . . substitution property of equality

∠BAE = ∠BCE . . . . addition property of equality

∠ABE = ∠CBE = 90° . . . . BE ⊥ AC

BE ≅ BE . . . . reflexive property of congruence

ΔBEA ≅ ΔBEC . . . . ASA congruence theorem

∠BAE ≅ ∠BCE . . . . CPCTC

9514 1404 393

Answer:

  B. a = plus-or-minus 4

Step-by-step explanation:

  3a² -21 = 27 . . . . . . . given

  3a² = 48 . . . . . . . . . . add 21 to both sides (desired form)

  a² = 16 . . . . . . . . . . .  divide both sides by 3

  a = ±4 . . . . . . . take the square root

Answer:

a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.

b) 0.0306 = 3.06% probability that all five students selected are males.

c) 0.0131 = 1.31% probability that all five students selected are females.

d) 0.9869 = 98.69% probability that at least one male is selected.

Step-by-step explanation:

The students are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this question:

15 + 13 = 28 students, which means that [tex]N = 28[/tex]

5 are selected, which means that [tex]n = 5[/tex]

13 females, which means that [tex]k = 13[/tex]

a. 3 females and 2 males are selected?

3 females, so this is P(X = 3).

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 3) = h(3,28,5,13) = \frac{C_{13,3}*C_{15,2}}{C_{28,5}} = 0.3056[/tex]

0.3056 = 30.56% probability that 3 females and 2 males are selected.

b.all five students selected are males?

0 females, so this is P(X = 0).

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 0) = h(0,28,5,13) = \frac{C_{13,0}*C_{15,5}}{C_{28,5}} = 0.0306[/tex]

0.0306 = 3.06% probability that all five students selected are males.

c. all five students selected are females?

This is P(X = 5). So

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 5) = h(5,28,5,13) = \frac{C_{13,5}*C_{15,0}}{C_{28,5}} = 0.0131[/tex]

0.0131 = 1.31% probability that all five students selected are females.

d.at least one male is selected?

Less than five females, so:

[tex]P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869[/tex]

0.9869 = 98.69% probability that at least one male is selected.