Solve the inequality: y+4<8

Hurry!! pls answer!! Rewrite this exponential equation as a logarithmic equation.

lara31 [8.8K]

It should be log_e(4)=x

7 0

2 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

2 years ago

i have been trying to solve this compound and double angle question please help me find the answer to these question guys

natali 33 [55]

Answer:

These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form, I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it" anyway,

Step-by-step explanation:

a)

next the question asks 1-cos 2A and this is total abuse of notation. the way this should be written is 1- cos( 2A) so we know that the A is part of the cosine functions input... btw.. in any computer program, it would never ever let you get away with that top form of the expression. :/ anyway... I keep ranting.. huh... sorry :P

1-cos(2A) is an odd form of the identity 1/2(1-cos(2A) = $sin^{2}$ (A) the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely, totally wrong, but I just miss one small thing :/ anyway....

our 1-cos(2A) needs the 1/2 added to it. or if we move that 1/2 to the other side it looks like 2* $sin^{2}$ (A) = 1-cos(2A) and this is that "odd" from of the identity that I was talking about.

next let's deal with sin(2A) it has an identity of 2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !

now our fraction looks like [2* $sin^{2}$ (A)] / 2 sin(A)cos(A)

so cancel out one of the sines

2*sin(A) / 2 cos(A)

cancel the 2s

Sin(A) / Cos(A) = Tan(A)

nice it worked out :P

b)

by the above that we just worked out, then

Tan(15) = Sin(15) / Cos(15)

I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of

Sin(15) = (√3 - 1) / 2√2

Cos(A) = (√3 + 1) / 2√2

you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)

back to our problem, the 2√2 will cancel out

then we have

Tan(15) = (√3 - 1) /(√3 + 1)

in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact. Most math professors like that form.

4 0

2 years ago

A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

jonny [76]

Answer:

2580 ft-lb

Step-by-step explanation:

Water leaks out of the bucket at a rate of $\frac{0.15 \mathrm{lb} / \mathrm{s}}{1.5 \mathrm{ft} / \mathrm{s}}=0.1 \mathrm{lb} / \mathrm{ft}$

Work done required to pull the bucket to the top of the well is given by integral

$W=\int_{a}^{b} F(x) dx$

Here, function $F(x)$ is the total weight of the bucket and water $x$ feet above the bottom of the well. That is,

$F(x)=4+(42-0.1 x)$

$=46-0.1x$

$a$ is the initial height and $b$ is the maximum height of well. That is,

$a=0 \text { and } b=60$

Find the work done as,

$W=\int_{a}^{b} F(x) d x$

$=\int_{0}^{60}(46-0.1 x) dx$

$&\left.=46x-0.05 x^{2}\right]_{0}^{60}$

$=(2760-180)-0[$

$=2580 \mathrm{ft}-\mathrm{lb}
$

Hence, the work done required to pull the bucket to the top of the well is $2580 \mathrm{ft}- \mathrm{lb}$

4 0

2 years ago

How do you simplify the square root of 48

noname [10]

When you simplify the square root of a number, you are looking for factors of the number that are perfect squares that you can remove from under the radical. For example sq root of 8... it's not a perfect square but factors of 8 include 2 x 4... 4 is a perfect square so we can take it out of the radical by taking it's square root. Now you are left with 2* sq root of 2.

For sq root of 48

48 = 12 x 4

12 = 3x4 so we can say 48 = 3x4x4

so... sq root 48 = sq root 3x4x4

we have two 4's under the radical. Sq root x^2 = x so sq root 4^2 is 4

Pull it out from the radical and we are left with 4 * sq root 3

4 0

2 years ago