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ryzh [129]
3 years ago
14

Find all the factors of 38,39 and 40 do they have any factors in common explain how you can tell if some numbers have factors in

commas without finding the factor
Mathematics
1 answer:
zvonat [6]3 years ago
6 0

Answer:

38= 1, 2, 19 and 38

39= 1,3, 13 and 39

40= 1,2,4,5,8,10,20 and 40

common factors is 1

to know the factors in commas without finding the factor is to simply divide the number by each factor gotten, take for example i have 38 as the number and i have gotten 1, 2.

38 divided by 1 equals 38

38 divided by 2 equals 19

so they are the factors

Step-by-step explanation:

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Answer:

x = 3.3

Step-by-step explanation:

A equation is given to us and we need to solve out for x. The given equation is ,

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a car salesperson earns a $500 fee per car she sells.if she sells 4 cars in one day,how much money does she earn in fee?
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Evaluate the dot product a⃗ ⋅b⃗ of a⃗ =3i^−8j^ and b⃗ =−2i^−6j^.
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Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evalua
Strike441 [17]

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

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

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

MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92

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

LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

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Then, the probability of getting a average score of 79 or higher is:

P(X>79)=P(z>-0.5)=0.69146

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

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e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

6 0
3 years ago
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