Can someone help? You don't have to give the direct answers though.

Bruce's hourly wage increased from $15.50 to $18.60. What ratr of increase does this represent?

Galina-37 [17]

Answer:

New price 3.10

Step-by-step explanation:

5 0

3 years ago

mina picks 63 flowers from her garden. She can put 10 flowers in each vase how many vases can she fill. how many extra flowers w

Lina20 [59]

Mina can fill 6 vases and she will have 3 left over.

6 0

3 years ago

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2xy and 5xy 7 and 2xy -3 and 5xy 7 and -3 Which are like terms

Tpy6a [65]

Answer:

2xy and 5xy are like terms

Step-by-step explanation:

8 0

3 years ago

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A teacher categorized students by whether they regularly completed their homework (or not) and whether they passed the course (o

ella [17]

Probabilities are used to determine the chances of an event.

Let P represents the event that a student passes the course, and C represents the event that a student completes the assignment

(a) The probability a student passed the course given that they completed their homework?

From the table, we have:

$\mathbf{n(P\ n\ C) = 25}$

$\mathbf{n(C) = 30}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P\ n\ C)}{n(C)}}$

This gives

$\mathbf{Pr = \frac{25}{30}}$

$\mathbf{Pr = 0.8333}$

Express as percentage

$\mathbf{Pr = 83.33\%}$

Hence, the probability a student passed the course given that they completed their homework is 83.33%

(b) The probability a student passed the course given that they didn't complete their homework

From the table, we have:

$\mathbf{n(P\ n\ C'= 7}$

$\mathbf{n(C') = 25}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P\ n\ C')}{n(C')}}$

This gives

$\mathbf{Pr = \frac{7}{25}}$

$\mathbf{Pr = 0.28}$

Express as percentage

$\mathbf{Pr = 28\%}$

Hence, the probability a student passed the course given that they didn't complete their homework is 28%

(c) The probability a student completed their homework given that passed the course?

From the table, we have:

$\mathbf{n(P\ n\ C) = 25}$

$\mathbf{n(P) = 32}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P\ n\ C)}{n(P)}}$

This gives

$\mathbf{Pr = \frac{25}{32}}$

$\mathbf{Pr = 0.78125}$

Express as percentage

$\mathbf{Pr = 78.125\%}$

Approximate

$\mathbf{Pr = 78.13\%}$

Hence, the probability a student completed their homework given that they passed the course is 78.13%

(d) The probability a student didn't complete their homework given that they didn't pass the course

From the table, we have:

$\mathbf{n(P'\ n\ C'= 18}$

$\mathbf{n(P) = 23}$

So, the required probability is:

$\mathbf{Pr = \frac{n(P'\ n\ C')}{n(P')}}$

This gives

$\mathbf{Pr = \frac{18}{23}}$

$\mathbf{Pr = 0.7826}$

Express as percentage

$\mathbf{Pr = 78.26\%}$

Hence, the probability a student didn't complete their homework given that they didn't pass the course is 78.26%