All categories

3 years ago

Your chance of winning a game is 4 out of 7. If you play the game 111 times, how many times would you expect to win?

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

Step-by-step explanation:

Using cross multiplication, you can set up this equation: $\frac{4}{7} = \frac{x}{111}$

You will get 444 = 7x, then divide each side by 7 to get a decimal. Rounded to the nearest whole-number, is 63.

You might be interested in

(») Find the missing height in the diagram.

ollegr [7]

Answer:

5 cm

Step-by-step explanation:

Multiply every number by 2. 3x2 = 9, 2x2 = 4, and 5x2 = 10

6 0

3 years ago

In a random sample of 25 ?people, the mean commute time to work was 30.4 minutes and the standard deviation was 7.1 minutes. ass

seropon [69]

In this problem, the given values are:

x = average = 30.4

s = standard deviation = 7.1

n = number of samples = 25

Degrees of freedom = n -1 =24

Using the t-distribution table for a normal curve at 90% CI, we get t-crit:

t-crit = 1.711

Now the Margin of Error (E) is calculated as:

E = t-crit*(s/ $\sqrt{n}$ )

By substituting the known values into the equation:

E = 1.711 * (7.1/ $\sqrt{8}$ )

E = +- 4.30

Therefore, the margin of error is 4.36. Hence, the commute time is between 26.1 minutes and 34.7 minutes.

7 0

3 years ago

***20 points “ if you answer

kykrilka [37]

I believe it is f... i hope i help

4 0

3 years ago

Non linear??????????

Artist 52 [7]

Answer:

Yes. That is correct

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the antiderivative of e^x^2?

ddd [48]

Answer: The antiderivative of $e^{x^{2}}$ is $\sqrt{\pi} \sqrt{e^{x^2}-1} +C$ .

Explanation:

$I=\int{e^{x^2}dx}\\I^2=\int{e^{x^2}dx}\int{e^{x^2}dx}$

Using dawson integral.

$I^2=\int{e^{x^2}dx}\int{e^{y^2}dy}$

$I^2=\int{\int{e^{x^2+y^2}dxdy}}$

Put $x^2+y^2=r^2$

$I^2=\int{\int{e^{r^2}rdrd\theta}}$

$\int_{0}^{2\pi}{d\theta}\int_{0}^{x}{re^{r^2}dr}$

Use substitution method and sunbstitute $r^2=t$ .

$\int_{0}^{2\pi}{d\theta}\int_{0}^{x^2}{\frac{1}{2}e^tdt}$

$I^2=\frac{1}{2}|\theta|_{0}^{2\pi}|e^t|_{0}^{x^2}\\I^2=\frac{1}{2}(2\pi)(e^{x^2}-1)\\I=\sqrt{\pi}\sqrt{e^{x^2}-1}$ .

Therefore, the antiderivative of $e^{x^{2}}$ is $\sqrt{\pi} \sqrt{e^{x^2}-1} +C$ .

4 0

3 years ago