What is the solution the the equation 1/2n+3=6

Y=4x-2, y=x+3 the solution to the system of equation

ella [17]

Y=4x-2=x+3
3x=5
x=5/3
y=5/3+3=14/3

6 0

3 years ago

Ethan has a box of assorted tea bags. There are seven green tea bags, six peppermint tea bags, five lemon tea bags, and six plan

Likurg_2 [28]

B. Randomly select and replace a tea bag 80 times. Then, observe how close to 20 times a peppermint tea bag is drawn.

3 0

3 years ago

Please help 20 points and brainliest

Maksim231197 [3]

Answer: the last second one: 3/4 0.8 94%

Step-by-step explanation:

because 3/4 is smaller than one so it has to go first.

hope it helps :) pls give brainliest

7 0

2 years ago

Read 2 more answers

Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to

polet [3.4K]

Answer:

(a) The probability that X is at most 30 is 0.9726.

(b) The probability that X is less than 30 is 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.

Step-by-step explanation:

We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.

Let X = the number among these that are nonconforming and can be reworked

The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).

Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.

Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).

So, the new mean of X, $\mu$ = $n \times p$ = $200 \times 0.11$ = 22

and the new standard deviation of X, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{200 \times 0.11 \times (1-0.11)}$

= 4.42

So, X ~ Normal( $\mu =22, \sigma^{2} = 4.42^{2}$ )

(a) The probability that X is at most 30 is given by = P(X < 30.5) {using continuity correction}

P(X < 30.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{30.5-22}{4.42}$ ) = P(Z < 1.92) = 0.9726

The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.

(b) The probability that X is less than 30 is given by = P(X $\leq$ 29.5) {using continuity correction}

P(X $\leq$ 29.5) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{29.5-22}{4.42}$ ) = P(Z $\leq$ 1.70) = 0.9554

The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is given by = P(15 $\leq$ X $\leq$ 25) = P(X < 25.5) - P(X $\leq$ 14.5) {using continuity correction}

P(X < 25.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{25.5-22}{4.42}$ ) = P(Z < 0.79) = 0.7852