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Julli [10]
3 years ago
7

Alex surveyed his classmates to determine who has been surfing and who has gone snowboarding let a be the event that a classmate

has snowboard it and let be be the event that a classmate has served if alex picked a classmate at random what is p(b|a)
Mathematics
1 answer:
strojnjashka [21]3 years ago
8 0

Answer: 3/4

Step-by-step explanation: Given that B is the event occurring and A is the event that has happened, we will use 36/48 as our probability. Once simplifying the numbers, you will get .75 or 3/4.

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Your mom opened a college fund for you when you were born. The fund had an annual interest rate of 2.5%. When you turned 18, the
Scilla [17]
Given:
Interest = 4,275
interest rate = 2.5%
term = 18 years
Principal = ?

Interest = Principal * interest rate * term
4,275 = P * 2.5% * 18
4,275/ (2.5% * 18) = P
4,275 / 0.45 = P
9,500 = P

Your mom deposited 9,500 when you were born.
3 0
3 years ago
If i have a 83.5 as my grade and let's just say i would get a 70/100 what would be my grade?
daser333 [38]

Answer:

76.75

Step-by-step explanation:

83.5 + 70=  153.5

153.5 / 2 = 76.75

Your grade would be 76.75.

HOPE THIS HELPED

6 0
3 years ago
Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =
marta [7]

Answer:

The integral for the arc of length is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

By using Simpon’s rule we get: 1.5355453

And using technology we get:  2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx

We need the derivative of the function:

f'(x)=\frac{1}{x}

And we need it squared:

[f'(x)]^2=\frac{1}{x^2}

Then the integral is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

Now, the Simposn’s rule with n=4 is:

\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)

In this problem:

a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}

So, the Simposn’s rule formula becomes:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then simplifying a bit:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then we just do those computations and we finally get the approximation via Simposn's rule:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%

8 0
3 years ago
Hi , the answer I choose was wrong , what is the correct answer ? Will choose brainlest so be quick
dimaraw [331]

Answer:

SAS

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Hii! please help i’ll give brainliest
natima [27]

Answer:

Bukhara

Hope this helps!

--Applepi101

6 0
3 years ago
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