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2 years ago

Jerry is a judge he hears 5 cases every 2 3/8 per hours Jerry hears case at a constant rate How many cases does he hear per hour

Mathematics

2 answers:

Nataliya [291]2 years ago

5 0

Around 2 I think... 2.10 is what I got

sveticcg [70]2 years ago

5 0

Answer:

$\frac{40}{19}\frac{cases}{hour}$ or $2\frac{2}{19}\frac{cases}{hour}$

Step-by-step explanation:

we know that

Jerry hears $5$ cases every $2\frac{3}{8}$ hours

To find how many cases does he hear per hour, divide $5$ cases by $2\frac{3}{8}$ hours

convert mixed number to an improper fraction first

$2\frac{3}{8}=\frac{2*8+3}{8}=\frac{19}{8}$

$\frac{5}{(19/8)} =\frac{40}{19}\frac{cases}{hour}$

convert to mixed number

$\frac{40}{19}\frac{cases}{hour}=\frac{38}{19}+\frac{2}{19}=2\frac{2}{19}\frac{cases}{hour}$

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Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and

Black_prince [1.1K]

Answer:

0.33411

The probability that a particular criminal trial lasted atleast 24 days is 0.33411

Step-by-step explanation:

The random variable X is normally distributed with a mean of 21 and standard deviation of 7

;

X ~ N(μ = 21 ; σ = 7)

X ~ N(21, 7)

Probability that a trial lasted atleast 24 days :

P(X ≥ 24) :

The standardized score :

Z = (x - μ) / σ ; (24 - 21) / 7 ; 3 / 7 = 0.4286

Hence,

P(Z ≥ 0.4286) = 0.33411

The probability that a particular criminal trial lasted atleast 24 days is 0.33411

6 0

2 years ago

These tables represent a quadratic function with a vertex at (0,3) What us the average rate of change from the interval from x=7

VladimirAG [237]

The rate of change is shown to the right of the second graph as being -2

At 5 to 6 the average rate of change is -11, so at 6 to 7 the rate of change would be -11-2 = -13, then from 7 to 8 it would be -13-2 = -15.

The answer is D. -15

3 0

2 years ago

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who

padilas [110]

Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

$P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568$

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

$P_{AA} = 0.13(0.47) = 0.0611$