All categories

3 years ago

Pls help with all of them

Mathematics

1 answer:

Irina-Kira [14]3 years ago

3 0

6h is the same thing as saying 6 times h. If we plug 6 in for h, we get 6x6=36. 6x6 does equal 36, so h = 6 is a solution to the problem.

If you plug in 60 for y, you get 60-45=15. 60-45 actually does equal 15, so both sides are equal and y=60 is a solution to the problem

If you plug in x=1/4, you get 1/4 + 1/2 = 3/4. 1/2 = 2/4, and 2/4 + 1/4 = 3/4 so x=1/4 is a solution to the problem

If you plug in w=2 for #5, you get 6x2=120. 6x2 equals 12, not 120, so w=2 is not a solution to the problem

If you plug in n=30 into #6, you get 30/3=12. 30/3 = 10, not 12, so n=30 is not a solution.

Hope this helps!

You might be interested in

I WILL GIVE BRAINLIEST

e-lub [12.9K]

The answer is 16. because the 4 lines up with the 16

8 0

3 years ago

Read 2 more answers

A 7th-grade class has 14 girls and 10 boys. An 8th-grade class has 11 girls and 13 boys. What is the probability of picking a st

melomori [17]

Answer:

14>10 11<13

Step-by-step explanation:

4 0

3 years ago

WILL MARK BRAINLIEST!!!

Ludmilka [50]

Answer:

(C. one solution

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

123456789

Margaret [11]

Use long subtraction to evaluate.
$418.51

3 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: