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3 years ago

Help, I need the area

Mathematics

1 answer:

GenaCL600 [577]3 years ago

6 0

Answer- 314.16

Area =3.14 x r^2

Area =3.14 x 10^2= 314.16

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Is the answer is c please help me and explain

Zolol [24]

Answer:

5/8.

Step-by-step explanation:

3/4 equals 6/8 and 6/8 minus 1/8 (being subtracted) is 1/8.

I did not use Siri

6 0

3 years ago

Problem 7.43 A chemical plant superintendent orders a process readjustment (namely shutdown and setting change) whenever the pH

vaieri [72.5K]

Complete Question

Problem 7.43

A chemical plant superintendent orders a process readjustment (namely shutdown and setting change) whenever the pH of the final product falls below 6.92 or above 7.08. The sample pH is normally distributed with unknown mu and standard deviation 0.08. Determine the probability:

(a)

of readjusting (that is, the probability that the measurement is not in the acceptable region) when the process is operating as intended and $\mu$ = 7.0 probability

(b)

of readjusting (that is, the probability that the measurement is not in the acceptable region) when the process is slightly off target, namely the mean pH is $\mu$ = 7.02

Answer:

The value is $P(X < 6.92 or X > 7.08 ) = 0.26431$

The value is $P(X < 6.92 or X > 7.08 ) = 0.29344$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 7.0$

The standard deviation is $\sigma = 0.08$

Considering question a

Generally the probability of readjusting when the process is operating as intended and mu 7.0 is mathematically represented as

$P(X < 6.92 or X > 7.08 ) = P(X < 6.92 ) + P(X > 7.08)$

=> $P(X < 6.92 or X > 7.08 ) = P(\frac{X - \mu }{\sigma} < \frac{6.9 - 7}{0.08} ) + P(\frac{X - \mu}{\sigma} > \frac{7.08 - 7}{0.08} )$

Generally

$\frac{X - \mu }{\sigma} = Z(The \ standardized \ value \ of \ X)$

=> $P(X < 6.92 or X > 7.08 ) = P(Z < \frac{6.9 - 7}{0.08} ) + P(Z > \frac{7.08 - 7}{0.08} )$

=> $P(X < 6.92 or X > 7.08 ) = P(Z < -1.25) + P(Z > 1 )$

From the z table the probability of (Z < -1.25) and (Z > 1 ) is

$P(Z < -1.25) = 0.10565$

and

$P(Z > 1 ) = 0.15866$

=> $P(X < 6.92 or X > 7.08 ) = 0.10565 + 0.15866$

=> $P(X < 6.92 or X > 7.08 ) = 0.26431$

Considering question b

Generally the probability of readjusting when the process is operating as intended and mu 7.02 is mathematically represented as

$P(X < 6.92 or X > 7.08 ) = P(X < 6.92 ) + P(X > 7.08)$

=> $P(X < 6.92 or X > 7.08 ) = P(\frac{X - \mu }{\sigma} < \frac{6.9 - 7.02}{0.08} ) + P(\frac{X - \mu}{\sigma} > \frac{7.08 - 7.02}{0.08} )$

Generally

$\frac{X - \mu }{\sigma} = Z(The \ standardized \ value \ of \ X)$

=> $P(X < 6.92 or X > 7.08 ) = P(Z < \frac{6.9 - 7.02}{0.08} ) + P(Z > \frac{7.08 - 7.02}{0.08} )$

=> $P(X < 6.92 or X > 7.08 ) = P(Z < -1.5) + P(Z > 0.75 )$

From the z table the probability of (Z < -1.5) and (Z > 0.75 ) is

$P(Z < -1.5) = 0.066807$

and

$P(Z > 0.75 ) = 0.22663$

=> $P(X < 6.92 or X > 7.08 ) = 0.066807 + 0.22663$

=> $P(X < 6.92 or X > 7.08 ) = 0.29344$

6 0

3 years ago

If p is the original price of a car, which equation best represents the situation

slava [35]

Answer:

p - 3000 = 16000

Step-by-step explanation:

none of the options given are correct

3 0

3 years ago

The arithmetic mean (A) of two numbers (a and b) is given by the formula A=a+b/2, and their geometric mean (G) is given by G=√ab

Ann [662]

A = a+b / 2
G = √( a b )
G = √ ( A H ) / ² ( we will square both sides of the equation )
G² = A H
H = G² / 2
$H = \frac{G ^{2} }{A}= \frac{a b}{A}= \\ = \frac{ab}{ \frac{a+b}{2} }= \frac{2ab}{a+b}$

6 0

4 years ago

Read 2 more answers

Construct the indicated confidence interval for the population mean using the t-distribution. Assume the population is normally

DanielleElmas [232]

Using the t-distribution, it is found that the confidence interval is (14.2, 14.8).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

t is the critical value.

n is the sample size.

s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a <em>two-tailed 95% confidence interval</em>, with 18 - 1 = <em>17 df</em>, is t = 2.1098.

The other parameters are given by:

$\overline{x} = 14.5, s = 0.68, n = 18$ .

Hence, the bounds of the interval are given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 14.5 - 2.1098\frac{0.68}{\sqrt{18}} = 14.2$

$\overline{x} + t\frac{s}{\sqrt{n}} = 14.5 + 2.1098\frac{0.68}{\sqrt{18}} = 14.8$

The confidence interval is (14.2, 14.8).

More can be learned about the t-distribution at brainly.com/question/16162795

4 0

3 years ago