All categories

3 years ago

Help........ASAP.............

Mathematics

2 answers:

sasho [114]3 years ago

6 0

3/5 because you have a probability of pulling 3 cards out of 5 cards

RoseWind [281]3 years ago

4 0

Answer:

3/5

Step-by-step explanation:

You might be interested in

Indefinite integral need help pleaseeeeeee

tatiyna

Integrate indefinite integral:
$I=\int\frac{dx}{e^{2x}+3e^x+2}dx$

Solution:
1. use substitution $u=e^x$
=>
$du=e^xdx$
=>
$dx=\frac{du}{e^x}$
=>
$dx=\frac{du}{u}$
=>
$I=\int\frac{du}{u(u^2+3u+2)}du$
$=\int\frac{du}{u(u+2)(u+1)}du$
2. decompose into partial fractions
$\frac{1}{u(u+2)(u+1)}$
$=\frac{A}{u}+\frac{B}{u+2}+\frac{C}{u+1}$
where A=1/2, B=1/2, C=-1
$=\frac{1}{2u}+\frac{1}{2(u+2)}-\frac{1}{u+1}$
3. Substitute partial fractions and continue
$I=\int\frac{du}{2u}+\int\frac{du}{2(u+2)}-\int\frac{du}{u+1}$
$=\frac{log(u)}{2}+\frac{log(u+2)}{2}-log(u+1)}$
4. back-substitute u=e^x
$=\frac{log(e^x)}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}$
$=\frac{x}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}$

Note: log(x) stands for natural log, and NOT log10(x)

7 0

3 years ago

6,000 dollars is invested in two different accounts earning 3% and 5% interest. At the end of one year, the two accounts earned

kap26 [50]

Answer:

$2000

Step-by-step explanation:

6,000 dollars is invested in two different accounts earning 3% and 5% interest.

Let x be the amount invested in account earning 3% interest

So, 6000-x is the amount invested in account earning 5% interest

$S.I. =\frac{P\times R\times T}{100}$

So, Simple interest for account earning 3% interest

$S.I. =\frac{x \times 3 \times 1}{100}$

$S.I. =0.03x$

Simple interest for account earning 5% interest

$S.I. =\frac{(6000-x) \times 5 \times 1}{100}$

$S.I. =0.05(6000-x)$

$S.I. =300-0.05x$

Since we are given that the two accounts earned $220 in interest

So, $0.03x+300-0.05x=220$

$−0.02x+300=220$

$80=0.02x$

$\frac{80}{0.02}=x$

$4000=x$

So, 4000 was invested at 3%

(6000-x)=6000-4000=2000 was invested at 5%

Hence 2000 was invested at 5%

8 0

4 years ago

Can somebody help me with this question?

Naya [18.7K]

Answer:

whats the question

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The prior probabilities for events A1 and A2 are P(A1) = 0.30 and P(A2) = 0.55. It is also known that P(A1 ∩ A2) = 0. Suppose P(

k0ka [10]

We're given the following probabilities:

$P(A_1)=0.30$

$P(A_2)=0.55$

$P(A_1\cap A_2)=0$

$P(B\mid A_1)=0.20$

$P(B\mid A_2)=0.05$

(a) Yes, $A_1$ and $A_2$ are mutually exclusive. This is exactly what zero probability of their intersection means. The two events cannot occur simultaneously.