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3 years ago

Consider this exponential equation:

Mathematics

1 answer:

ioda3 years ago

4 0

Step-by-step explanation:

F( x ) = 5 (3) ^ x

Y intercept

Let x = 0

f(0) = 5× 3 ^ 0

f(0 = 5 × 1

f(0) = 5

X intercept

let y = o

0 = 5 × 3 ^x

No x intercept/ zero

therefore

Vertical intercept (0; 5)

Domain XER

▪︎this refer to the values of X

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What is the greatest common factor GCF of 10 and four

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Take the numbers 50 and 30. Their greatest common factor is 10, since 10 is the greatest factor that both numbers have in common. To find the GCF of greater numbers, you can factor each number to find their prime factors, identify the prime factors they have in common, and then multiply those together.

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The nth term of a sequence is 3n + 2

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Answer:

see explanation

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To calculate the first 3 terms substitute n = 1, 2, 3 into the n th term rule

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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.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5$

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.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

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.

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