Ask question

Ask question

All categories

3 years ago

14

4/5 of all the orange juice in the world is produced in Brazil what percent of all the orange juice in the world is produced in

Brazil

1 answer:

klasskru [66]3 years ago

3 0

The answer is 80% For the question

You might be interested in

Plzz how do i find the area w just the base

Roman55 [17]

Answer:

8x8= 64 divided by 2 = 37

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

One type of insect is 0.0052 meters long. What is the lenght in scientific notation

Ket [755]

Answer:

5.2 × $10^{-3}$

Step-by-step explanation:

a number in scientific notation is expressed as

a × $10^{n}$

where 1 ≤ a < 10 and n is an integer

write 0.0052 as a number between 1 and 10, that is 5.2

we have to the move the decimal point 3 places to the left to retain it's value

3 places left means n is - 3

0.0052 = 5.2 × $10^{-3}$

8 0

3 years ago

6% greater than or less than 48%

elena-s [515]

6% is less than 48%. Due to the reason that 48% is a larger portion.

3 0

3 years ago

Read 2 more answers

Jessica has a shelf that is 26 inches long . If she cuts off 12 3/8 inches what is the new length of the shelf ?

maw [93]

14 and 5/8 inches long now

5 0

3 years ago

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:

$\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}$ .

Therefore:

$\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}$ .

Lookup $0.1985$ on a normal distribution table. The corresponding $z$ -score is $0.52$ . (In other words, $P(0 \le Z \le 0.52) = 0.1985$ .)

Given that

$z = 0.52$ ,
$x =5$ , and
$\sigma = 2$ ,

Solve the equation $\displaystyle z = \frac{x - \mu}{\sigma}$ for the mean, $\mu$ :

$\displaystyle 0.52 = \frac{5 - \mu}{2}$ .

$\mu = 5 - 2 \times 0.52 = 3.96$ .

3 0

3 years ago