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Damm [24]
3 years ago
7

PLS HELP MEH ;-; I will be giving brainliest for who ever answers right

Mathematics
2 answers:
expeople1 [14]3 years ago
7 0
In line one she forgot to multiply 4 on the left side by -1

320=30+15s
290=15s
S=19.33
S=19
19 student tickets.
Anettt [7]3 years ago
3 0
I think cause she multiple wrong
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Which of the following numbers is divisible by 9? <br><br>A.) 111<br>B.) 222<br>C.) 333<br>D.) 444
Leto [7]

Answer:

c

Step-by-step explanation:

3+3+3=

6 0
3 years ago
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(8-(2-2))∙9<br><br> 9<br><br> 34<br><br> 72<br><br> 118
dsp73

Answer: 72

Step-by-step explanation:

(8-(2-2))×9

First do (2-2=0)

(8-0=8)

8×9=72

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3 years ago
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3. Adult train tickets cost $25, student tickets costs $15 and child tickets cost $10. If a group of a
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Step-by-step explanation:

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3 years ago
Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or impo
Rudik [331]

Answer:

a) Probability that vehicle is Light Truck is 0.69

b) Probability that vehicle is imported car is 0.08 or 8%.

Step-by-step explanation:

Part a)

Event that vehicle is a car is given by A. Since, the vehicles can only be classified as cars or light trucks, the event that a vehicle will be light truck will be compliment of A.

i.e. Event that vehicle is a Light Truck = A^{c}

It is given that in recent year 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.

So, from here we can say that, if a vehicle is randomly selected from all the vehicles the probability that it would be a light truck will be:

P(Vehicle will be Light Truck) = P(A^{c}) = 69% = 0.69

Part b)

Event that vehicle is imported is given by B. We need to find the probability the a randomly chosen vehicle is an imported car i.e. we have to find probability of occurrence of events A and B together, which will be denoted as: P(A and B)

Since, P(A^{c}) = 69% = 0.69

P(A) = 1 - P(A^{c}) = 1 - 0.69 = 0.31

It is also given that 78% of vehicles were domestic. This means, the percentage/probability of imported vehicles is:

P(B) = 1 - 0.78 = 0.22

55% of vehicles were domestic light trucks. This can be expressed as:

P(A^{c}\&B^{c}), the compliment of this event will give us P(A or B).

i.e.

P(A or B) = 1 - P(A^{c} \&B^{c}) = 1 - 0.55 = 0.45

According to the addition rule of probabilities:

P(A or B) = P(A) + P(B) - P(A and B)

Substituting the calculated values gives us:

0.45 = 0.31 + 0.22 - P(A and B)

P(A and B) = 0.08

This means, the probability that vehicle is imported car is 0.08 or 8%.

4 0
3 years ago
Please solve this question
xxTIMURxx [149]

Answer:

I think its C

Step-by-step explanation:

3 0
3 years ago
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