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exis [7]
3 years ago
12

Solve the system of equations

Mathematics
1 answer:
nekit [7.7K]3 years ago
8 0

Answer: x = 3 and y = 1

Step-by-step explanation:

y = - x + 4 ............... equation 1

y = x - 2 ............. equation 2

Solving the linear equation by substitution method , we have :

substitute equation 2 into equation 1 , that is

x - 2 = - x + 4

Add 2 to both sides , we have

x = -x + 4 + 2

Add x to both sides , we have

2x = 6

divide through by 2

x = 3

Substitute x = 3 into equation 2 to find the value of y , we have

y = x - 2

y = 3 - 2

y = 1

Therefore : x = 3 and y = 1

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A model ocean liner is 12 inches long. If it is built to a scale of 1 in:300 ft., how long is the real ocean liner?
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3600 ft

Step-by-step explanation:

because each inch in the model is 300 feet in real life, you do 12 times 300 ft and you get 3600 ft

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Assume that​ women's heights are normally distributed with a mean given by mu equals 61.4 in​, and a standard deviation given by
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3 years ago
F(x) = 4x2 – 10x + 2<br> Find f(-7)
eimsori [14]

Answer:

f(-7) = 268

Step-by-step explanation:

f(x) = 4x^2 – 10x + 2

Let x = -7

f(-7) = 4(-7)^2 – 10(-7) + 2

     Exponents first

f(-7) = 4(49) – 10(-7) + 2

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8 0
3 years ago
Read 2 more answers
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.

6 0
3 years ago
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