I used to describe a food item having a spicy or salty quality without being sweet?

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the

Aleks [24]

Answer:

Check the explanation

Step-by-step explanation:

Ans=

A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134

M = 10: 1 – 0.9973^10 = 0.0267

M = 20: 1 – 0.9973^20 = 0.0526

M = 30: 1 – 0.9973^30 = 0.0779

M = 50: 1 – 0.9973^50 = 0.126

18)

Ans=

Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values

7 0

3 years ago

Solve for a 16 = - 8a

Dima020 [189]

Hello! The answer to this question is: A = -2

3 0

3 years ago

Factor the expression completely. 63x – 9

liraira [26]

Answer:

The answer is 9(7x – 1).

Step-by-step explanation:

63x – 9

3 × 3 × 7 × x – 3 × 3

9(7x – 1)

Thus, The answer is 9(7x – 1).

-TheUnknownScientist 72

4 0

2 years ago

Freeeeeeeeeeeeeeeeeeeeeee points

LuckyWell [14K]

Answer:

heyyy

Step-by-step explanation:

heyyyy im just here to help in getting them pts because im smart like that :)

7 0

2 years ago

Read 2 more answers

An automobile manufacturer claims that its van has a 31.3 miles/gallon (MPG) rating. An independent testing firm has been contra

Yuliya22 [10]

Answer:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

$z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82$

$p_v =2*P(z$

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is NOT significantly different from 31.3 at 2% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=31.1$ represent the sample mean

$\sigma=1.3$ represent the population standard deviation for the sample

$n=140$ sample size

$\mu_o =31.3$ represent the value that we want to test

$\alpha=0.02$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We can assume that we conduct a hypothesis in order to check if the true mean is equal to 31.3 , the system of hypothesis would be:

Null hypothesis: $\mu =31.3$

Alternative hypothesis: $\mu \neq 31.3$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82$

P-value

Since is a two-sided test the p value would be:

$p_v =2*P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is NOT significantly different from 31.3 at 2% of signficance.

5 0

3 years ago

I used to describe a food item having a spicy or salty quality without being sweet?​

I used to describe a food item having a spicy or salty quality without being sweet?