Find LCD of 5/8x, 7/4x+8

Rick, Maya, and Tom work in a call center. On a particular day, Rick worked 5 and ½ hours, Maya worked 6.1 hours, and Tom worked

storchak [24]

5h 30m + 6h 6m + 6h 30m = 17 hours and 66 minutes or 18 hours and 6 minutes.

1,110 minutes total.

Hope I helped! :)

5 0

4 years ago

2(x-1)=3(x-4) how many solutions

34kurt

Distribute
2(x-1) =3(x-4)
2x -2 =3 (x-4)
Distribute
2x-2 =3 (x-4)
2x-2=3x-12
Add 2 both sides of the equation
2x-2=3x-12
2x-2+2=3x-12+2
Simplify
2x=3x-10
Subtract 3x from both sides of the equation
2x=3x -10
2x -3x =3x -10 -3x
Simplify
-x=-10
Divide both sides of the equation by the same term
-x=-10
-x/-1 / -1 / -10/-1
Simplify
X=10
Answer : x=10

6 0

3 years ago

Read 2 more answers

A store has a $159.98 item on sale for $79.99, plus an additional 20% off. What is the percentage of savings off on this item?

Mama L [17]

The answer would be 70 percent

6 0

4 years ago

Read 2 more answers

The weight W that a horizontal beam can support varies inversely as the length L of the beam. Suppose that an 8 dash m beam can

miss Akunina [59]

<h2>Answer:</h2>

The number of kilograms a 19 m beam can support is:

378.9474 kg

<h2>Step-by-step explanation:</h2>

It is given that:

The weight W that a horizontal beam can support varies inversely as the length L of the beam.

Let k denote the constant.

This means that the weight W and Length L is related as:

$W=\dfrac{k}{L}$

Suppose that an 8 dash m beam can support 900 kg.

This means that L=8 and W=900

Then,

$900=\dfrac{k}{8}\\\\i.e.\\\\k=900\times 8\\\\i.e.\\\\k=7200$

Now , we are asked to find the value of W when L= 19

i.e.

$W=\dfrac{7200}{19}\\\\i.e.\\\\W=378.9474\ kg$

6 0

3 years ago

A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probab

stira [4]

Answer:

$P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}$

Step-by-step explanation:

Hi!

Lets define:

C = 1 if candidate is qualified

C = 0 if candidate is not qualified

A = 1 correct answer

A = 0 wrong answer

T = 1 test passed

T = 0 test failed

We know that:

$P(C=1)=q\\P(A=1 | C=1) = p\\P(A=0 | C=0) = p$

The test consist of 20 questions. The answers are indpendent, then the number of correct answers X has a binomial distribution (conditional on the candidate qualification):

$P(X=x | C=1)=f_1(x)=\binom{20}{x}p^x(1-p)^{20-x}\\P(X=x | C=0)=f_0(x)=\binom{20}{x}(1-p)^xp^{20-x}$

The probability of at least 15 (P(T=1))correct answers is:

$P(X\geq 15|C=1)=\sum_{i=15}^{20}f_1(i)\\P(X\geq 15|C=0)=\sum_{i=15}^{20}f_0(i)\\$

We need to calculate the conditional probabiliy P(C=1 |T=1). We use Bayes theorem:

$P(C=1|T=1)=\frac{P(T=1|C=1)P(C=1)}{P(T=1)}\\P(T=1) = qP(T=1|C=1) + (1-q)P(T=1|C=0)$

$P(T=1)=q\sum_{i=15}^{20}f_1(i) + (1-q)\sum_{i=15}^{20}f_0(i)\\P(T=1)=\sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i)]$

$P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}$

5 0

3 years ago