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2 years ago

Over a three-week period, Abram spent 15

Mathematics

1 answer:

denis23 [38]2 years ago

7 0

Answer:

20 hours

Step-by-step explanation:

First off, if we divide the number of hours he has played by the number of weeks 15/3 =

It would be an average of 5 hours per week

Therefore, in the next 4 weeks we would practice for 20 more hours.

i hope this helps you<33

have a nice day or night

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How to multiple common factors plsss help

Juliette [100K]

Answer:

We can find the common multiples of two or more numbers by listing the multiples of each number and then finding their common multiples. For example, to find the common multiples of 3 and 4, we list their multiples and then find their common multiples. Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,

Step-by-step explanation:

this is the rai d right?

3 0

3 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

3 years ago

Which strategy is the most appropriate strategy to solve 5x^2+30x ?

guapka [62]

B. Square root property

4 0

3 years ago

Write the equations included in the same set of related facts as 6×8 =48

choli [55]

I dont think this is right but (2x3)x(2x4)=48 sorry i dont think that was right

8 0

3 years ago

Read 2 more answers

The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determ

steposvetlana [31]

Solution:

From Bernoulli equation

$\frac{1}{2}$ ρ $V_{1}^{2}$ + ρgh = $\frac{1}{2}$ ρ $V_{2}^{2}$ , where ρ is density of water, h – height difference and $V_{1}$ and $V_{2}$ are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

$\pi r_{1}^{2}$ $V_{1}$ = $\pi r_{2}^{2}$ $V_{2}$ , where $A_{1}$ and $A_{2}$ are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that $V_{1} = \frac{V_{2}}{4}$

Inserting it back into Bernoulli equations produces:

$V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s$ and the flow rate is

$\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}$

6 0

3 years ago