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Roger is having a picnic for 78guests. He plans to serve each guest at least one hot dog. If each package, p, contains eight hot

german

Question is Incomplete; Complete question is given below;

Roger is having a picnic for 78 guests. He plans to serve each guest at least one hot dog. If each package, p, contains eight hot dogs, which inequality could be used to determine how many packages of hot dogs Roger will need to buy?

1) $p \geq 78$

2) $8p \geq 78$

3) $8 +p \geq 78$

4) $78 + p \geq 8$

Answer:

2) $8p \geq 78$

Step-by-step explanation:

Given:

Number of guest in the picnic = 78 guest

Number of hot dog each guest will have = 1

Number of hot dogs in each package = 8 hot dogs.

We need to write the In equality used to determine the number of packages of hot dogs roger must buy

Solution:

Let the number of packages be 'p'.

First we will find the total number of hot dogs required.

so we can say that;

total number of hot dogs required is equal Number of guest in the picnic multiplied by Number of hot dog each guest will have.

framing in equation form we get;

total number of hot dogs required = $78\times 1 =78$

Now we can say that;

Number of hot dogs in each package multiplied by number of packages should be greater than or equal to total number of hot dogs required.

framing in equation form we get;

$8p\geq 78$

Hence The In equality used to determine the number of packages of hot dogs roger must buy is $8p\geq 78$ .

6 0

3 years ago

The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago

Max and his family visited the beach

Alex17521 [72]

They stayed at the beach for 7 days. 4 x 3 = 12. 12 - 5 = 7

7 0

3 years ago

Read 2 more answers

Carey creates the table below to help him determine 75% of 36.

Setler [38]

Answer:

27

Step-by-step explanation:

75×36=2700

2700/100=27

7 0

3 years ago

Is 3 a factor of 73? ( Yes ) or ( No )

Mkey [24]

Answer:

The correct answer is no.

Step-by-step explanation:

If a number is a factor of another number, that means that it can be equally divided into it evenly.

I know that 3 x 20 = 60, and 73 - 60 = 13.

Since 3 is not a factor of 13, it is also not a factor of 73.

Hope this helps! :D

6 0

2 years ago