Ask question

Ask question

All categories

2 years ago

6

A 5 gallon mixture contains 40% acid. A 3 gallon mixture contains 50% acid. What percent is obtained by putting the two mixture

s together?

1 answer:

Nady [450]2 years ago

6 0

(5/8)40 + (3/8)50 = 25 + 18.75 = 43.75%

You might be interested in

Whats the equation to find the total cost ?

Bezzdna [24]

Y = 8x
Is the equation

7 0

2 years ago

1. For a 30 student speech course, in how many ways can the student be selected to give the first 5 speeches?

liraira [26]

Combination problem:

1) 30 C 5 = 142506

2) 13 C 4 = 715

8 0

3 years ago

18 ➗ (-4 1/2) with explanation

slamgirl [31]

Answer:

-4

Step-by-step explanation:

If anything the first thing you would want to do is covert your fraction so a decimal. (-4 1/2) as a decimal is -4.50. From there it would be easier to divide. to get an answer of -4

8 0

2 years ago

Can anyone help me please

Lena [83]

Answer:

$55.24

Step-by-step explanation:

Add $32.78 + $27.24 + $5.22

= 65.24

Subtract 10

= $55.24

4 0

2 years ago

Read 2 more answers

A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a)

marta [7]

(a) If $A,B,C$ are mutually exclusive, then

$P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0$

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

$P(A\cap B)=P(A)P(B),$

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

$P(A\cap B\cap C)=P(A)P(B)P(C)$

so that

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(A\cap B)+P(B\cap C))+P(A\cap B\cap C)$

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

$P(A\cup B\cup C)=\dfrac35$

5 0

3 years ago