The probability of an event NOT happening.

Branden and Pete both play running back. Branden carries the ball 75 times for 550 yards and Pete had 42 carries for 380 yards w

mars1129 [50]

Answer:

Pete runs farther per carry.

Step-by-step explanation:

Given:

Branden and Pete both play running back.

Branden carries the ball 75 times for 550 yards.

Pete had 42 carries for 380 yards

Question asked:

Who runs farther per carry ?

Solution:

First of all we find each boy's distance covered per carry:-

For Branden

Branden carries the ball 75 times for = 550 yards.

Branden carries the ball 1 time for = $\frac{550}{75} =7.34\ yard$

For Pete

Pete carries 42 times for = 380 yards

Pete carries 1 time for = $\frac{380}{42} =9.05\ yards$

As Pete runs 9.05 yards for 1 carry while Branden runs 7.34 yards for 1 carry, hence Pete runs farther per carry.

3 0

3 years ago

You withdraw $29.79 from your bank account. Now your balance is −$20.51. Write and solve an equation to fi nd the amount of mone

Lunna [17]

You had $9.28 prior to making the withdrawal

-$20.51 = x - $29.79
x = $9.28

4 0

3 years ago

Read 2 more answers

A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou

Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question, $x = 5$ and $\sigma = 2$ . The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$ .

To solve for $\mu$ , the mean of this distribution, the only thing that needs to be found is the value of $z$ . Since

The problem stated that $P(X \le 5) = 69.85\% = 0.6985$ . Hence, $P(Z \le z) = 0.6985$ .

The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$ . To estimate $z$ from $P(Z \le z) = 0.6985$ , it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$ , $z > 0$ . As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$ . Besides, $P(Z < 0) = P(Z \le 0) = 0.50$ . As a result: