The cost of 40 candy bars is

Del dinero que tenía gasté 1/3 en comida y 1/8 en pasaje, quedándome aún 39 pesos. ¿Cuánto tenía al principio?

ruslelena [56]

Answer: 72 pesos.

Step-by-step explanation:

Supongamos que al principio tenias X, donde X representa la cantidad de dinero que tenias en un principio.

Gastaste:

X/3 en comida

X/8 en pasaje.

Entonces el gasto total es:

X/3 + X/8 = X*(1/3 + 1/8) = X*(11/24)

Entonces el dinero que te queda es:

X - X*(11/24) = X*(13/24) = 39 pesos.

X = (24/13)*39 pesos = 72 pesos

Al principio tenias 72 pesos.

6 0

4 years ago

All athletes at the Olympic Games (OG) are tested for performance-enhancing steroid drug use. The basic Anabolic Steroid Test (A

Semenov [28]

Answer:

a ) the probability of using steroids and having a negative test is 0.5%

b) The probability of testing positive is 6.4%

c) The probability of not using steroids, given that the test is negative is 99.47%

d) No, they are not statistically indepent.

e) The probability that the athlete will either use steroids or test positive is 6.9%

Step-by-step explanation:

Let A be the event that the test result is positive and B the event that the athlete uses Steroids. We are given the following

$P(A|B) = 90%, P(A|B^c) = 2%, P(B) = 5%$

From which we deduce that

$P(A^c|B) = 10%, P(B^c) = 95%$

a) We are asked for the probability $P(A^c\cap B)$ . REcall the conditional probability formula that, given two events C,D the conditional probability $P(C|D) = \frac{P(C\cap D)}{P(D)}$ . Then we have that

$P(A^c\cap B) = P(A^c|B)P(B) = 10\% \cdot 5\%=0.5\%$ .

b) We are asked for the probability P(A). We can use the fact that given two mutually exclusive events(that is, whose intersection is empty) A,B the probability P(C) of an event is given by $P(C) = P(C|A)P(A)+P(C|B)P(B)$ . Then

$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = 90\%\cdot5\% + 2\% \cdot 95% = 6.4\%$

c) We are asked for the probability P(B^c|A^c). Recall that $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . Then

$P(B^c|A^c) = \frac{P(A^c|B^c)P(B^c)}{P(A^c)}= \frac{P(A^c|B^c)P(B^c)}{1-P(A)}= \frac{(1-P(A|B^c))P(B^c)}{1-P(A)}=\frac{98\%\cdot 95\%}{1-6.4\%}= 99.47\%$

d) We say that two events A,B are statistically indepent if P(A|B) = P(A). Note that from point B the probability of testing negative is 1- 6.4% = 93.6%. Since 93.6% is different from 99.47% this means that testing positive and using steroids are not statistically independent.

e) We are asked for the probability $P(A\cup B)$ . We use the following

$P(A\cupB) = P(A)+P(B)-P(A\cap B) = P(A) +P(B)-P(A|B)P(B) = 6.4\%+5\%-90\%\cdot 5\%=6.9\%$

4 0

4 years ago

Find the value of f(-2) for the function f\left(x\right)=5x^2-4

Flauer [41]

Answer:

f(- 2) = 16

Step-by-step explanation:

Substitute x = - 2 into f(x), that is

f(- 2) = 5(- 2)² - 4 = 5(4) - 4 = 20 - 4 = 16

6 0

3 years ago

Read 2 more answers

Convert 50 percentage into fraction

Paul [167]

Answer:

50/100, simplified to 1/2

Step-by-step explanation:

50% means half of 1.

1/2=.5 which is half of 1.

1=100%

.5=50%

1/2=50%

6 0

3 years ago

Read 2 more answers

Which graph represents the given equation?<br> y<br> =<br> NICO<br> + 4r – 2

guajiro [1.7K]

Equations can be represented on graphs by plotting the corresponding x and y values. The graph of $y = \frac{3}{2}x^2 + 4x - 2$ is option (B).