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3 years ago

a chef bought 8 packs of hot dog buns with a total of 48 buns. what is the unit rate for the number in 1 pack? please help.

Mathematics

2 answers:

Romashka [77]3 years ago

7 0

There are 6 buns in each pack.
:D

tamaranim1 [39]3 years ago

4 0

You would just do 48/8 which equals 6

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3 years ago

Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passe

satela [25.4K]

Answer:

The probability of SFS and SSF are same, i.e. P (SFS) = P (SSF) = 0.1311.

Step-by-step explanation:

The probability of a component passing the test is, P (S) = 0.79.

The probability that a component fails the test is, P (F) = 1 - 0.79 = 0.21.

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Compute the probability of the test result as SFS as follows:

P (SFS) = P (S) × P (F) × P (S)

$=0.79\times0.21\times0.79\\=0.131061\\\approx0.1311$

Compute the probability of the test result as SSF as follows:

P (SSF) = P (S) × P (S) × P (F)

$=0.79\times0.79\times0.21\\=0.131061\\\approx0.1311$

Thus, the probability of SFS and SSF are same, i.e. P (SFS) = P (SSF) = 0.1311.

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4 years ago

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3 years ago

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Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.

adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

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3 years ago