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2 years ago

I need help ive been stuck on these questions

Mathematics

1 answer:

Alina [70]2 years ago

5 0

I think its a hope it helps

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the sum of two numbers is 123. when the larger number is added to 5 times the smaller, the sum is 343. find the numbers.

vovikov84 [41]

Create a system of equations
x + y = 123
5x + y = 343

I use substitution
y = 123 - x
5x + 123 - x = 343
4x + 123 = 343
4x = 220
x = 55

Plug in
x + y = 123
55 + y = 123
y = 68

Check
5x + y = 343
5(55) + 68 = 343
275 + 68 = 343
343 = 343

3 0

1 year ago

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What is the percentage of 2.47 out of 100

quester [9]

2.47 is 2.47% of 100

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2 years ago

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Suppose p(a) = 0.40 and p(a 

sveticcg [70]

Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

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

For parts (a) and (b), you're given everything you need to determine $P(B)$ .

For part (c), if $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ . If the given probability is $P(A\cup B)=0.55$ , then you can find $P(B)=0.15$ . But if this given probability is for the intersection, finding $P(B)$ is impossible.

For part (d), if $A$ and $B$ are independent, then $P(A\cap B)=P(A)\cdot P(B)$ .

8 0

3 years ago

PLS HELP WILL MARK BRAINLIEST

gavmur [86]

Answer:

1. Rolling a number less than 5

The numbers on a standard dice are 1, 2, 3, 4, 5, and 6. Therefore it is possible to roll a 6, but it is much more likely to be 1-5, simply because there are more of them.

2. Rolling a 2

Again, it is more likely that the result will be one of the other 5 numbers. But obviously a 2 is still an option, so it's not impossible.

3. Rolling a number less than 10

Every option is less than 10 so it is certain.

4 0

3 years ago

Beth is going to the State Fair which costs $10 to get in. Each ride costs $5.50. If he has $45 to spend at the Fair, what is th

Jet001 [13]

Answer:same2434..[-

Step-by-step explanation:25790=/,svxy

5 0

2 years ago

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I need help ive been stuck on these questions​

I need help ive been stuck on these questions