I saw two of these and they are both from you so i’m gonna try to help.. the last question on the last page is 4. I’m so sorry. I don’t know anything else. I tried looking it up but nothing worked i’m sorry. I just wish i could help and do more..
Answer:
r = 21
Step-by-step explanation:
The "one step" is to undo the division by 3. To undo that division, you multiply both sides of the equation by 3. this gives you ...
r·(3/3) = 7·3
r = 21
_____
The basic idea of any equation solution process is to "undo" what has been done to the variable. This is where the multiplication and addition identity elements come into play.
In this problem, the variable is multiplied by 1/3. The number that you multiply this by to get the identity element for multiplication (1) is the inverse of this fraction: 3/1, or 3. That is, 3 × 1/3 = 3/3 = 1. When 1 multiplies r, the result is just r, which is what you're trying to get to.
The rules of equality tell you that whatever you do to one side of an equation, you must also do to the other side. If you multiply r/3 by 3 on the left, then you must also multiply 7 by 3 on the right to keep the equation a true statement.
_____
<em>If the "one-step" involves addition ...</em>
If the equation had been ...
r + 3 = 7
Then the operation you need to "undo" is the addition of 3. That is accomplished by adding -3 to both sides of the equation. Then you have ...
r + 3 - 3 = 7 - 3
r + 0 = 4
r = 4
You will recognize 0 as the additive identity element: r + 0 = r. In order to get that (0) as a sum, you need to add opposites: +3 -3 = 0. Again, you have to do the same thing (add -3) to both sides of the equation in order to keep it true.
Answer:
Probability that fewer than 2 of these parts are defective is 0.604.
Step-by-step explanation:
We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.
A random sample of 7 parts produced by this machine is chosen.
The above situation can be represented through Binomial distribution;
where, n = number of trials (samples) taken = 7 parts
r = number of success = fewer than 2
p = probability of success which in our question is % of defective
parts produced by one of the machine, i.e; 19%
<em>LET X = Number of parts that are defective</em>
<u>So, it means X ~ Binom(n = 7, p = 0.19)</u>
Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)
P(X < 2) = P(X = 0) + P(X = 1)
= 
= 
= <u>0.604</u>
<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>
8+2>4 ig that would be the answer..
