I’ll mark as BRANLIEST!!!

stepan [7]

75.25x + 3604 = 300.50x
300.50x - 75.25x = 3604
225.25x = 3604
x = 16
Both companies charged the same at 16 tons
first answer 16 tons
--------------------
300.50x = 300.50(16) = $4,808
both companies charged the same are $4,808
second answer is $4,808

5 0

4 years ago

What percentage increase when 8 is made to 10? by step

KATRIN_1 [288]

Answer:

25% increase

Step-by-step explanation:

4 0

3 years ago

Geometry question for homework.

mars1129 [50]

So remember that the distance formula is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

1.

$\sqrt{(4-4)^2+(-2-(-5))^2}\\ \sqrt{0^2+3^2}\\ \sqrt{0+9}\\ \sqrt{9}\\ 3$

Distance between (4,-5) and (4,-2) is 3 units.

2.

$\sqrt{(1-(-1))^2+(1-4)^2}\\ \sqrt{2^2+(-3)^2}\\ \sqrt{4+9}\\ \sqrt{13}$

Distance between (1,1) and (-1,4) is √13 (or 3.61 rounded to the hundreths) units.

3 0

3 years ago

How can 25/9 be represented as a decimal? PLeasehelp! 1st answer is brainleuses

dolphi86 [110]

Answer:

2.8 or 2.77777

Step-by-step explanation:

9 can go into 25 less than 3 times but more than 2

5 0

3 years ago

Read 2 more answers

Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p

Ymorist [56]

Answer:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

$A\sim Uniform(1 ,7)$

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

$B\sim Exp(\lambda=\frac{1}{30 min})$

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

$E(A +B)$

Since the two random variables are assumed independent, then we have this

$E(A+B) = E(A)+E(B)$

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got: