All categories

2 years ago

a rectangle is 5 feet longer than it is wide. the perimeter of the rectangle is 34 feet. what is the length of the rectangle?

2 answers:

4 0

P = 2(L + W)...perimeter of a rectangle formula
P = 34
L = W + 5

34 = 2(W + 5 + W)
34 = 2(2W + 5)
34 = 4W + 10
34 - 10 = 4W
24 = 4W
24/4 = W
6 = W.....width is 6 ft

L = W + 5
L = 6 + 5
L = 11 <=== length is 11 ft

DochEvi [55]2 years ago

3 0

The length is 11 feet and the width is 7 feet long

You might be interested in

Deposits $10 in an account that earns an interest rate of 5% compounded annually. What is the balance in the account at the end

d1i1m1o1n [39]

<span>Compound Interest is calculated on the initial payment and also on the interest of previous periods.

So after 1 year w/ 5% interest, $10 -> $10*1.05 = $10.5
after 2 year, the total becomes $10.5*1.05 = $11.025

formula to calculate is 10*(1.05)^50 = $114.67

ans is A</span>

8 0

2 years ago

Identify the graph of the solution set of : x^2 -9/ x^2+6x+8

Snowcat [4.5K]

Answer:

The answer is the 1st one

Step-by-step explanation:

I risked my answer lucky for you i got it right :)

5 0

3 years ago

Read 2 more answers

a number is divided by 3 and the result is added to 1. the new result is doubled to give the final result 80. whats the number

Gnesinka [82]

78, because you have to do the inverse operation so 80/2= 40 then 40-1= 39 and 39*3 = 78

6 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

2 years ago

A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters

PilotLPTM [1.2K]

Answer:

e. 0.0072

Step-by-step explanation:

We are given that a bottling company uses a filling machine to fill plastic bottles with cola. And the contents vary according to a Normal distribution with Mean, μ = 298 ml and Standard deviation, σ = 3 ml .

Let Z = $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1) where, Xbar = mean contents of six randomly

selected bottles

n = sample size i.e. 6

So, Probability that the mean contents of six randomly selected bottles is less than 295 ml is given by, P(Xbar < 295)

P(Xbar < 295) = P( $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{295 - 298}{\frac{3}{\sqrt{6} } }$ ) = P(Z < -2.45) = P(Z > 2.45)

Now, using z% score table we find that P(Z > 2.45) = 0.00715 ≈ 0.0072 .

Therefore, option e is correct .

7 0

3 years ago