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Ilya [14]
3 years ago
13

if the sides of a square are lengthened by 3 m, the area becomes 81 m2. Find the length of a side of the original square.

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
4 0

Answer:

The length of a side of the original square is 6 meters

Step-by-step explanation:

Let

x ----> the length side of the original square

The area of a square is equal to

A=b^2

where

b is the length side of the square

In this problem

The new length side of the square is b=(x+3)\ m and the area is A=81\ m^2

so

(x+3)^2=81

solve for x

take square root both sides

(x+3)=9

x=9-3=6\ m

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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

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x = (A1 - E1*A1) / B1

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3 years ago
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3/11 divided by 3/11
vladimir2022 [97]

Answer: 1

Step-by-step explanation:

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Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

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b) p=0.04

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P(x>1)= 1 - 0.368 - 0.442=0.190

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In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

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