Solve for x 5*2^3x=21

You roll a number cube and flip a coin.What is the probability of rolling a number less than 2 and flipping tails. Write your an

vesna_86 [32]

Answer:

1/6

Step-by-step explanation:

number cube 6 faces less than 2, only 2 possibilities, leaving us with a fraction of 2/6=1/3

tails is a 50 percent chance or 1/2

3/6 time 2/6

equals:

1/6

the probability of rolling a number less than 2 and flipping tails on a coin is 1/6

find probability of event a: number cube and multiply it by probability of event b: coin

8 0

3 years ago

Stephen wrote out the following steps to compute the difference quotient, v(x) = x3 + 8.

pogonyaev

Answer:

3x2 + 3xh + h2

Step-by-step explanation:

c

8 0

2 years ago

Find the area between the following curves. y=x^3-x^2+x+8 ; y=5x^2-7x+8

solmaris [256]

Find where the two curves meet.

$x^3 - x^2 + x + 8 = 5x^2 - 7x + 8 \\\\ \implies x^3 - 6x^2 + 8x = 0 \\\\ \implies x (x-2) (x - 4)= 0 \implies x=0, x=2, x=4$

The area between the curves is

$\displaystyle \int_0^4 \left|\left(x^3-x^2+x+8\right) - \left(5x^2 - 7x + 8\right)\right| \, dx = \int_0^4 \left|x(x-2)(x-4)\right| \, dx$

When $x$ is between 0 and 2, $x(x-2)(x-4)$ is positive; when $x$ is between 2 and 4, $x(x-2)(x-4)$ is negative. So we split the integral at $x=2$ to get

$\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_2^4 x(x-2)(x-4)\,dx$

In the second integral, substitute $y=x-2$ to get

$\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_0^2 (y+2)y(y-2)\,dy$

$\displaystyle \int_0^2 x(x-2) \bigg((x-4) - (x+2)\bigg) \, dx$

$\displaystyle -6 \int_0^2 x(x-2) \, dx$

$\displaystyle 6 \int_0^2 \left(2x - x^2\right) \, dx$

$\displaystyle 6 \left(x^2 - \frac13x^3\right)\bigg|_0^2 = \boxed{8}$

3 0

1 year ago

Vehicles entering an intersection from the east are equally likely to turn left, turn right, or proceed straight ahead. If 50 ve

ira [324]

Answer:

The probability that at least two-third of vehicles in the sample turn is 0.4207.

Step-by-step explanation:

Let X = number of vehicles that turn left or right.

The proportion of the vehicles that turn is, p = 2/3.

The nest n = 50 vehicles entering this intersection from the east, is observed.

Any vehicle taking a turn is independent of others.

The random variable X follows a Binomial distribution with parameters n = 50 and p = 2/3.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=50\times \frac{2}{3}=33.333>10\\\\n(1-p)=50\times \frac{1}{3}= = 16.667>10$

Thus, a Normal approximation to binomial can be applied.

So, $X\sim N(np, np(1-p))$

Compute the probability that at least two-third of vehicles in the sample turn as follows:

$P(X\geq \frac{2}{3}\times 50)=P(X\geq 33.333)=P(X\geq 34)$

$=P(\frac{X-\mu}{\sigma}>\frac{34-33.333}{\sqrt{50\times \frac{2}{3}\times\frac {1}{3}}})$

$=P(Z>0.20)\\=1-P(Z$

Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.

6 0

2 years ago

Your teacher is giving away candy from a bag. The bag contains 5 Reese's Peanut

scoray [572]

Answer:

P(Snickers n Snickers) = $\frac{2}{51}$

Step-by-step explanation:

Given

Bag of candy

5 Reese's Peanut Butter Cups,

4 Snickers

3 Sour Patch

6 Skittles

Required

Calculate P(Snickers n Snickers)

To solve this question, it'll be assumed that you and your friend selected snickers at a consecutive time.

First, we need to calculate the total number of candy

Total = 5 + 4 + 3 + 6

Total = 18 candies

P(Snickers n Snickers) means P(Snickers and Snickers);

i.e. the probability of selecting snicker twice

Number of snickers = 4;

So, the probability of you selecting a snicker first is $\frac{4}{18}$

This is gotten by dividing number of snickers by total candies.

At this point, there are 4 - 1 snickers left and there are 18 - 1 candies left;

So, the probability of your friend selecting a snicker is $\frac{4 - 1}{18 - 1}$ .

So, P(Snickers n Snickers) = $\frac{4}{18}$ * $\frac{4 - 1}{18 - 1}$

P(Snickers n Snickers) = $\frac{4}{18}$ * $\frac{3}{17}$

P(Snickers n Snickers) = $\frac{2}{9}$ * $\frac{3}{17}$

P(Snickers n Snickers) = $\frac{2}{3}$ * $\frac{1}{17}$

P(Snickers n Snickers) = $\frac{2}{51}$

Hence, the probability of you and your friend choosing snickers is $\frac{2}{51}$

4 0

3 years ago