Combine these radicals. √27- √3

Abdullah is a quality control expert at a factory that paints car parts. He knows that 20\ , percent of parts have an error in t

-Dominant- [34]

The error rate has decreased after changing the painting process.

<u>Step-by-step explanation:</u>

Abdulla knows that 20 percent of the parts have an error in their painting. After suggesting changes in painting process, he wants to know whether the error rate has changed.

Number of parts in the random sample=400400400

Number of parts that had an error=606060

We have to determine what percentage of 400400400 is 606060

$606060=x/100 \times 400400400\\=0.15%$

After changing the painting process 0.15% of parts have error.

The previous percentage was 20.Hence the error rate has clearly changed.

3 0

2 years ago

In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t

victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is $p =\frac{1}{5}$

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5

b

The mean is $\mu = 20$

The standard deviation is $\sigma=4$

c

The probability, $P =0.0002$

d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability $Pe = 0.0003$

The probabilities $P \approx Pe$

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

$p =\frac{1}{5}$

$=0.2$

The mean of this distribution is mathematical represented as

$\mu = np$

substituting the value

$\mu = 100 * 0.2$

$\mu = 20$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np (1-p)}$

substituting the value

$= \sqrt{100 * 0,2 (1-0.2)}$

$\sigma=4$

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

$P=P(X \ge 34 )$

By standardizing the normal approximation we have that

$P(X \ge 34) \approx P(Z \ge z)$

Now z is mathematically evaluated as

$z = \frac{x-\mu}{\sigma }$

Substituting values

$z = \frac{34-20}{4}$

$=3.5$

So using the z table the $P(Z \ge 3.5)$ is 0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

6 0

2 years ago

The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 inches. Heights of men are normally dist

Maksim231197 [3]

Answer:b)0.8577

Step-by-step explanation:

Since the heights of men are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = heights of men

u = mean height

s = standard deviation

From the information given,

u = 69 inches

s = 2.8 inches

We want to find the probability that the mean height of the 100 men is less than 72 inches.. It is expressed as

P(x < 72)

For x = 72

z = (72 - 69)/2.8 = 1.07

Looking at the normal distribution table, the probability corresponding to the z score is 0.8577

P(x < 72) = 0.8577

7 0

2 years ago

Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage ea

garik1379 [7]

Answer:

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

Step-by-step explanation:

We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.

We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.26.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26*0.74}{160}}\\\\\\ \sigma_p=\sqrt{0.0012}=0.0347$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0347=0.068$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.26-0.068=0.192\\\\UL=p+z \cdot \sigma_p = 0.26+0.068=0.328$

The 95% confidence interval for the population proportion is (0.192, 0.328).

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

3 0

2 years ago

At the beginning of year 1, paolo invests $500

suter [353]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annuall, thus once}
\end{array}\to &1\\
t=years\to &t
\end{cases}
\\\\\\
A=500\left(1+\frac{0.04}{1}\right)^{1\cdot t}\implies A=500(1+0.04)^t
\\\\\\
\textit{after 5 years }t=5\qquad A=500(1.04)^5$

the example on your picture uses A(n) and n = years, but is pretty much the same, in this case is t = years.

3 0

2 years ago

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