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3 years ago

I need help with number 35.

Mathematics

1 answer:

Sergio039 [100]3 years ago

6 0

Since ED and DB are exactly half of EB, we can set ED and DB equal to each other.

x + 4 = 3x - 8

Subtract x from both sides. Add 8 to both sides

12 = 2x

Divide both sides by 2

6 = x

ED = 6 + 4 = 10

ED = DB = 10

EB = ED +DB = 10 + 10 = 20

ED = 10

DB = 10

EB = 20

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The original price of a car is entered into spreadsheet cell A1 and the annual depreciation amount in cell B1.

Reil [10]

Answer:

A1 / B1 ;

(D1 - A1) / B1 ;

(A1 - E1*A1) / B1

Step-by-step explanation:

A1 = original price of car

B1 = annual Depreciation amount

Number of years it will take for the car to depreciate totally :

Using the straight line Depreciation relation :

y = mx + c

c = intercept = initial or original value of car

m = annual Depreciation amount

x = number of years

y = value after x years

For total Depreciation, final value, y = 0

0 = mx + c

mx = - c

x = - c / m

Hence, x = A1 / B1

B.)

D1 = car value

Length it will take for car to depreciate to value in D1 :

y = mx + c

y = D1; m = B1 ; c = A1

D1 = B1x + A1

B1x = D1 - A1

x = (D1 - A1) / B1

C.)

E1 = decrease percentage

Time it takes for car to decrease by percentage in E1

y = E1 * A1

E1 * A1 = B1x + A1

(A1 - E1*A1) = B1x

x = (A1 - E1*A1) / B1

4 0

3 years ago

40 points please help me due in 50 minutes!

velikii [3]

U use sin
sin50 = b/10
b=10(0.76)
b=7.6

3 0

3 years ago

WhaThe factorization of x2 + 3x – 4 is modeled with algebra tiles.t is the factored form of x2 – x – 2?

Nataly [62]

Answer:

x^2 -x -2 = (x -2)(x +1)

Step-by-step explanation:

To factor x^2 -x -2, you look for factors of -2 that have a sum of -1:

-2 = (-1)(2) = (1)(-2) . . . . . the latter pair has a sum of -1

Then these are the constants in the binomial factors:

x^2 -x -2 = (x +1)(x -2)

_____

<em>Comment on the question</em>

The fact that x^2 +3x -4 is modeled with algebra tiles seems to be irrelevant to the question that is actually asked.

4 0

3 years ago

List below are the speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, CA. Use the sample data to construct

zheka24 [161]

Answer:

The 99% confidence interval of the population standard deviation is 1.7047 < σ < 7.485

Step-by-step explanation:

Confidence interval of standard deviation is given as follows;

$\sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}$

s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$

Where:

$\bar x$ = Sample mean

s = Sample standard deviation

n = Sample size = 7

χ = Chi squared value at the given confidence level

$\bar x$ = ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857

The sample standard deviation s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$ = 2.854

The test statistic, derived through computation, = ±3.707

Which gives;

$C. I. = 57.857 \pm 3.707 \times \dfrac{2.854}{\sqrt{7} }$

$\sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{16.812}^{}}}< \sigma < \sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{0.872}}$

1.7047 < σ < 7.485

The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.

7 0

3 years ago

A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote <em>R</em> as the event that a component needs rework.

It is given that:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

6 0

3 years ago