Help me please! Will mark brainlist

Help please??!?? i don’t understandddddd

Zina [86]

Answer:

I think the answer is - $\frac{u}{6}$

Step-by-step explanation:

thats the answer i got in math.way

3 0

3 years ago

How do you find the least common for 3 and 8?

sesenic [268]

The least common multiple of 3 and 8 is 24.

The least common multiple of two numbers is the lowest value number that is a multiple of both the numbers. To find the least common multiple of 3 and 8, you first list their multiples:

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24...

Multiples of 8: 8, 16, 24, 32, 40, 48, 56...

When you compare the two lists of multiples, you can see that the number 24 is the lowest number that is a multiple of both 3 and 8, so it is their LCM, or least common multiple.

<em>HOPE THAT HELPS!!!!</em>

8 0

4 years ago

What is the first step needed to solve 2 over 5 multiplied by x minus 6 equals negative 16?

ololo11 [35]

<span>B.)Add 6 to both sides</span>

8 0

3 years ago

Read 2 more answers

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

(c) $P(X > 980) = 0.00001\\\\$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$