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If 70g is 100%.... 48g is taken away... what is the percentage of the remaining 22g. Rounded to the nearest whole number?

Ivenika [448]

we are given with 70 grams and is said that 48 grams is taken away. This is equal to 22 grams. To find the mass percentage, we divide the remaining mass and divide by the total mass. The remaining 22 grams is equal to 22/70 * 100 percent equal to 31.43 percent.

8 0

3 years ago

What is 0.6 as a fraction in simplest form

Illusion [34]

we know that

step 1

multiply $0.6$ by $\frac{10}{10}$

so

$0.6*\frac{10}{10} =\frac{6}{10}$

step 2

Divide numerator and denominator by $2$

$\frac{\frac{6}{2}}{\frac{10}{2}} =\frac{3}{5}$

therefore

the answer is

$\frac{3}{5}$

4 0

3 years ago

Read 2 more answers

How many degress are in an angle that cuts 2/4 of a circle

saw5 [17]

B:180 degrees because a circle is 360 degrees and half of 360 is 180. I'm learning this too lol

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:

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

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

write each equation in a slope-intercept form. 1. (6, 2) and (-3, -6) 2. (-4 and -6) and (3, -8) 3.(-2, -8) and (6, 4) 4.(-3, 9)

ANTONII [103]

Answer:

1. (6, 2) and (-3, -6) 2. (-4 and -6) and (3, -8) 3.(-2, -8) and (6, 4) 4.(-3, 9) and (0, 1) 5.(2, -2) and (8, 1) 6.(-12, 4) and (0, 2) 7.(-8, -4) and (4, -7) 8.(-4, 13) and (-2, 6)

Step-by-step explanation:

5 0

3 years ago