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2 years ago

How much can Sophia spend on party favors

Mathematics

1 answer:

Fantom [35]2 years ago

8 0

Think about you have a pizza. you have 6 people and a total of 12 slices. How many can each person get? 2.

This is the same thing. $2.47 divided by 6 people = $0.46 cents each

You might be interested in

there are six black socks, 6 blue socks, 12 white socks, and 4 pattern socks in the drawer. what is the probability of pulling a

Pepsi [2]

Answer:

3/98

Step-by-step explanation:

six black socks, 6 blue socks, 12 white socks, and 4 pattern socks= 28 socks

P( blue ) = blue /total = 6/28 = 3/14

Replace it

six black socks, 6 blue socks, 12 white socks, and 4 pattern socks= 28 socks

P ( pattern) = pattern/total = 4/28 = 1/7

P( blue,replace, pattern) = 3/14*1/7 = 3/98

7 0

2 years ago

Please see photo below for question!!

Liono4ka [1.6K]

No need for the diagram (though it helps)

P(AUB) = P(A) + P(B) - P(A and B),

P(A) = 0.61 - 0.41 + 0.18 = 0.38

So, D!!

4 0

3 years ago

Translate the sentence into an equation.

Vilka [71]

3(x - 8) = 4
3(x) - 3(8) = 4
3x - 24 = 4
+ 24 + 24
3x = 28
3 3
x = 9¹/₃

6 0

2 years ago

The probability that a student correctly answers on the first try (the event

Mademuasel [1]

Note that the two events are mutually exclusive. If the question is answered correctly on the first try, there's no need to give it another attempt. So $\mathbb P(A\cap B)=0$ .

We're given that $P(A)=0.2$ and $P(B\mid A^C)=0.5$ . From the first probability, we know that $P(A^C)=1-0.2=0.8$ . By definition of conditional probability,

$\mathbb P(B\mid A^C)=\dfrac{\mathbb P(B\cap A^C)}{\mathbb P(A^C)}$
$\implies\mathbb P(B\cap A^C)=0.5\cdot0.8=0.4$

We're interested in the probability of either $A$ or $B$ occurring, i.e. $\mathbb P(A\cup B)$ . Apply the inclusion-exclusion principle, which says

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

We know the probability of intersection is 0, and we know $\mathbb P(A)$ . Meanwhile, by the law of total probability, we have

$\mathbb P(B)=\mathbb P(B\cap A)+\mathbb P(B\cap A^C)=\mathbb P(B\cap A^C)$

so we end up with

$\mathbb P(A\cup B)=0.2+0.4=0.6$

3 0

3 years ago

the boggs family paid $49.75 for their car rental. How many miles did they drive? i need help please :')

ANTONII [103]

Hello.

49.75 = 30 + 0.05m

Subtract 30 from both sides.

49.75 - 30

19,75 = 0.05m

Divide both sides by 0.05

395 miles = m

Have a nice day

8 0

2 years ago