Factor the following expression 14x + 49

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 = <u><em>the number among these that are nonconforming and can be reworked</em></u>

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) = <u>0.9726</u>

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) = <u>0.9554</u>

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

P(X $\leq$ 14.5) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{14.5-22}{4.42}$ ) = P(Z $\leq$ -1.70) = 1 - P(Z < 1.70)

= 1 - 0.9554 = 0.0446

The above probability is calculated by looking at the value of x = 0.79 and x = 1.70 in the z table which has an area of 0.7852 and 0.9554.

Therefore, P(15 $\leq$ X $\leq$ 25) = 0.7852 - 0.0446 = 0.7406.

5 0

3 years ago

Verify that fand g are inverse functions (both questions please)

Grace [21]

Answer:

hope it helps you..,.....,..........

3 0

2 years ago

The information given below was extracted from the accounting records of Sunray Traders, a partnership business with Sunny and R

Crank

The preparation of the Profit and Loss Appropriation Account for the year ended February 28, 2022, is as follows:

<h3>What is the Profit and Loss Appropriation Account?</h3>

A profit and loss appropriation account is a special account prepared for the distribution of the partnership's profit and loss for the period.

For Sunray Traders Partnership, the profit and loss appropriation account is detailed below.

<h3>Profit and Loss Appropriation Account</h3>

For the year ended February 28, 2022

Sunny Raymond Total

Net Profit for the year R900,000

Interest on capital R84,000 R24,000 (R108,000)

Annual Salaries R307,500 R246,000 (R553,500)

Share of profits R119,250 R119,250 (R238,500)

<h3>Question Completion:</h3>

(c) The partners are entitled to the following monthly salaries from 01 March 2021 to 30 November 2021:

Sunny R25 000 Raymond R20 000

Note: The salaries of the partners were increased by 10% with effect from 01 December 2021.

d) The balance of the profit and loss must be shared equally between Sunny and Raymond.

Prepare the Profit and Loss Appropriation Account for the year ended February 28m 2022.

<h3>Data and Calculations:</h3>

DEBIT CREDIT

Capital: Sunny R1 000 000

Capital: Raymond 600 000

Current a/c: Sunny (01 March 2021) 40 000

Current a/c: Raymond (01 March 2021) 60 000

Drawings: Sunny R300 000

Drawings: Raymond 400 000

Net profit = R900 000

Sunny's Salaries = R307,500 (R25,000 x 9 + R25,000 x 1.1 x 3)

Raymond Salaries = R246,000 (R20,000 x 9 + R20,000 x 1.1 x 3)

<h3>Capital Accounts Balances:</h3>

Sunny Raymond Total

Capital accounts R1 000 000 R600 000 R1 600 000

Drawings accounts 300 000 400 000 700 000

Net Capital accounts R700 000 R200 000 R900 000

Interest on capital R 84 000 R24 000 R108 000

Thus, the profit and loss of Sunray Traders Partnership are shared equally as agreed.

Learn more about partnership accounts at brainly.com/question/22693552

#SPJ1

3 0

1 year ago

I need help fast please!!

kaheart [24]

Solution set is: ( $7,\frac{13}{3}$ )

Option B is correct

Step-by-step explanation:

We need to find the solution to system of equations

$y=\frac{1}{3}x+2\\y=\frac{4}{3}x-5$

Let:

$y=\frac{1}{3}x+2\,\,\,eq(1)\\y=\frac{4}{3}x-5\,\,\,eq(2)$

Putting value of y from eq(2) into eq(1)

$\frac{4}{3}x-5=\frac{1}{3}x+2\\Combining\,\,like\,\,terms:\\\frac{4}{3}x-\frac{1}{3}x=2+5\\Taking\,\,LCM\\\frac{4x-1x}{3}=7\\\frac{3x}{3}=7\\x=7$

So, value of x=7

Putting value of x into equation 1 to find the value of y:

$y=\frac{1}{3}x+2\\y=\frac{1}{3}(7)+2\\y=\frac{7}{3}+2\\Taking\,\,LCM\\y=\frac{7+2*3}{3}\\y=\frac{7+6}{3}\\y=\frac{13}{3}$

So, value of y = $\frac{13}{3}$

Solution set is: ( $7,\frac{13}{3}$ )

Option B is correct

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

3 0

3 years ago

On a road trip in Motanna, USA, Elise sees a road sign that says Helena 87. Elise tests the accuracy of her cars odometer and tr

mote1985 [20]

Answer:

No i dont think so but im not sure

Step-by-step explanation:

8 0

3 years ago

Factor the following expression 14x + 49​

Factor the following expression 14x + 49