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Talja [164]
3 years ago
5

Booker owns 85 vidoe games he has 3 shelves to put the games on.each shelf can hold 40 video games.how many more vidoe games doe

s he have room for
Mathematics
1 answer:
Ksju [112]3 years ago
7 0
He has 3 shelves...each shelf can hold 40 video games.....40 * 3 = 120...so he has room for 120 video games.

and if he already has 85, then he would need : (120 - 85) = 35 more to fill up the shelves
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Please help thank you ahead of time
vlabodo [156]
Okay so basically, i’m pretty sure you have to make them catch up with each other, if i’m correct), so if it’s 30 to each student plus 2500 then 2 students is 60 bucks, 3 students is 90 and so on so forth......then if 1 student - 60 then 2 is 120 and vice versa ya know. So keep adding on the amount due for each student plus the extra amount which doesn’t change and make them catch up with each other. Keep on adding student after student and adding the fee until they match up. Sorry it’s not the exact answer but i hope this explains it... if i’m right
6 0
2 years ago
James has 21 cups of flour he will pour the flour equally into 4 container how many cups of flour will be in each container
Degger [83]

Answer:

5 1/4 cups of flour will be in each container

Step-by-step explanation:

21 ÷ 4= 5.25

5.25= 5 1/4

hope this helps babe <3

8 0
3 years ago
A cylinder has diameter of 4ft and a height of 9ft. Explain whether halving the diameter has the same effect on the surface area
elena55 [62]

Answer:

See Below

Step-by-step explanation:

The surface area of cylinder is given by the formula:

SA=2\pi r^2 + 2\pi r h

Where

r is radius ( diameter is 4, so radius is 4/2 = 2)

h is height ( h = 9)

Lets find original surface are:

SA=2\pi r^2 + 2\pi r h\\SA=2\pi (2)^2 + 2\pi (2) (9)\\SA=8\pi +36\pi\\SA=44\pi

<u>Halving diameter:</u>

diameter would be 4/2 = 2, so radius would be 2/2 = 1

So, SA would be:

SA=2\pi r^2 + 2\pi r h\\SA=2\pi (1)^2 + 2\pi (1) (9)\\SA=2\pi +18\pi\\SA=20\pi

<u>Halving height:</u>

Height is 9, halving would make it 9/2 = 4.5

Now, calculating new SA:

SA=2\pi r^2 + 2\pi r h\\SA=2\pi (2)^2 + 2\pi (2) (4.5)\\SA=8\pi + 18\pi\\SA= 26\pi

Original SA is 44\pi,

Halving diameter makes it 20\pi

Halving height makes it 26\pi

So, halving diameter does not have same effect as halving height.

5 0
3 years ago
The rule below shows how the cost of renting a kayak depends on how long it is rented. multiply the number of hours by $16, then
allochka39001 [22]
C=16h-10. 16 times hours minus 10
3 0
3 years ago
For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.
nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  1-P(A)+1-P(B)\\\\-P(A\cap B)\leq  1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

Proof:

If S is a universal set then:

A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\

B can be  expressed as:

B=B\cap(A\cup A^{c})\\

If B is intersection of two disjoint sets then

P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)

Then (1) becomes

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

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})

where A and A ∩ Bc  are disjoint.

P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})

From axiom P(E)≥0

P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)

Therefore,

P(A)≥P(B)

8 0
3 years ago
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