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3 years ago

PLEASE HELP ALMOST DUE BRANILIEST AND 5 STARTS AND A THANKS PLSLSLSLSL

Mathematics

2 answers:

Arada [10]3 years ago

7 0

Answer:

m=3

Step-by-step explanation:

18=5m+3

-5m=3-18

=-15

m=15÷5

Luden [163]3 years ago

6 0

Answer:

m=3

Step-by-step explanation:

18=5m+3

15=5m

m=3

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I NEED HELP PLEASE ASAP!

Virty [35]

Answer:

make a line with arrows at the end. On the line, make a hash mark and label it -13. on the hash mark, make a filled-in circle and color the number line to the left of the circle, including the arrow.

Step-by-step explanation:

This is how you show it. :)

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4 years ago

Henry and Jose are gaining weight for football. Henry weighs 205 pounds and is gaining 2 pounds per week. Jose weighs 195 pounds

Digiron [165]

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3 years ago

The following data summarizes results from 939 pedestrian deaths that were caused by accidents . If one of the pedestrian deaths

olganol [36]

Answer:

Step-by-step explanation:

Hello!

The contingency table is attached.

The total of pedestrians is 909, not 939, there are 30 accidents less in the given data.

I've calculated the asked probabilities using a total of 909.

1. The probability that the pedestrian was intoxicated or the driver was intoxicated.

Two events are mutually exclusive when the occurrence of one prevents the occurrence of the other in one single performance of the experiment. (i.e there is no intersection between the events P(A∩B)=0)

When two events are mutually exclusive, the probability of the union of both elements is equal to the sum of their individual probabilities P(A∪B)= P(A) + P(B).

When the events aren't mutually exclusive, the probability of their union is equal to the summary of their probabilities minus the probability of the intersection between these two events P(A∪B)= P(A) + P(B) - P(A∩B)

Where:

∪ is the union of both events, in colloquial language represents "or"

∩ is the union of both events, or intersection of the events, in colloquial language represents "and"

In this case P(P₁ ∪ D₁) = P(P₁) + P(D₁) - P(P₁ ∩ D₁) = $\frac{306}{909} + \frac{132}{909} - \frac{81}{909}$ = 0.393

P₁ and D₁ are not mutually exclusive

2. The probability of the pedestrian was intoxicated or the driver was not intoxicated.

These two events aren't mutually exclusive.

P(P₁ ∪ D₂)= P(P₁) + P(D₂) - P(P₁ ∩ D₂)= $\frac{306}{909} +\frac{777}{909} *-\frac{225}{909}$ = 0.944

I hope it helps!

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3 years ago

Trains headed for destination A arrive at the train station at 15-minute intervals starting at 7:00 am, whereas trains headed fo

maks197457 [2]

Answer:

The answer to the question is 0.43. Let's explain why.

Step-by-step explanation:

A train arrives at the station at 15-minute intervals starting from 7:00 am which makes the train arrives at;

7:00

7:15

7:30

7:45

8:00

B train arrive at 12-minute intervals starting at 7:07 means B train arrives at;

7:07

7:19

7:31

7:43

7:55

Now, let's review the chances of a passenger arrives at the station any time between 7:00 am and 8:00 am

If the passenger arrives between 7:00 - 7:07 the passenger will get into B train which gives us the $\frac{7}{60}\\$ chance since an hour is 60 minutes and the passenger waited for 7 minutes.
If the passanger arrives between 7:07 - 7:15 passenger will get into A train which gives $\frac{8}{60}$ chance. If we apply this to the time limits it goes by;
Between 7:15 - 7:19 passenger will go to B destination which gives $\frac{4}{60}$ chance
Between 7:19 - 7:30 passenger will go to A destination means $\frac{11}{60}$ chance
Between 7:30 - 7:43 passenger will go to B destination means $\frac{13}{60}$ chance
Between 7:43 - 7:45 passenger will go to A destination which gives $\frac{2}{60}$ chance
Between 7:45 - 7:55 passenger will go to B destination means $\frac{10}{60}$ chance
An lastly between 7:55 - 8:00 passenger will go to A destination which gives us $\frac{5}{60}$ chance