1/4 is the same thing as 25%. So we know that we have to find 25% of 2000 pounds, which is 1 ton. Convert 25% into a decimal and then multiply.
2000 times .25 is 500.
One fourth of one ton is 500 pounds.
Answer:
20%
Step-by-step explanation:
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Answer:
The correct answer is 218 math textbooks and 259 sociology textbooks.
Step-by-step explanation:
To solve this problem, we can make a system of equations. Let the number of sociology textbooks sold be represented by the variable "s" and the number of math textbooks sold be represented by the variable "m". Using these variables, we can make two equations:
s + m = 477
m + 41 = s
There are many ways to solve this system of equations. One approach we can take is substituting the value for s given by the second equation into the first equation. This is modeled below.
s + m = 477
(m + 41) + m = 477
Combining like terms on the left side of the equation yields:
2m + 41 = 477
Subtracting 41 from both sides of the equation gives us:
2m = 436
Finally, dividing both sides of the equation by 2 gives us:
m = 218
To solve for the number of sociology textbooks, we can substitute into either of our original equations.
m + 41 = s
(218) + 41 = s
s = 259
Therefore, your answer is m = 218 and s = 259, or 218 math textbooks and 259 sociology textbooks were sold.
Hope this helps!
Answer:
0.6,0.7,0.3 neither disjoint nor independent.
Step-by-step explanation:
Given that at a large university, 60% of the students have a Visa card and 40% of the students have a MasterCard.
A= visa card
B = Master card
P(A) = 0.60 and P(B) = 0.40
P(AUB)' = 0.30
i.e. P(AUB) = 0.70
Or P(A)+P(B)-P(AB) =0.70
P(AB)= 0.30
Randomly select a student from the university.
1) the probability that this student does not have a MasterCard.

2. the probability that this student has either a Visa card or a MasterCard.
=
3. Calculate the probability that this student has neither a Visa card nor a MasterCard.
=
4. Are the events A and B disjoint? Are the events A and B independent?
A and B have common prob 0.30 hence not disjoint.
P(AB) ≠P(A)P(B)
Hence not independent
The first graph with the month in numbers because the X value which is the month never appears again while the other appears again or attached to-other y value