Solve each system of linear equations by substitution

When you substitute a value for x or y, you are x or y with a value.

Hitman42 [59]

Step-by-step explanation:

See whenever we have two variables like x and y.

And also we have given two equations in terms of x and y.

Example: ax + by = p and cx + dy = q

In order to get the solution, we need to find the value of x and y using the given two equations.

So here what we can do is:

Consider any one equation and get x or y from that equation.

i.e consider ax + by = p or x = $\frac{1}{a}*(p-by)$

Now plug value of x (or y) from the equation you have considered into the another equation.

i.e c( $\frac{1}{a}*(p-by)$ )+dy = q

Now solve this equation to get y (or x).

and after that we can find the other variable.

4 0

2 years ago

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Is 1000 a sequence in 50 65 80 95

SCORPION-xisa [38]

Answer: yes

Step-by-step explanation:

it’s the 41st term

5 0

1 year ago

The mean height of women in a country (ages 20-29) is 64 4 inches A random sample of 50 women in this age group is selected What

Simora [160]

Answer:

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 64.4 inches, standard deviation of 2.91

This means that $\mu = 64.4, \sigma = 2.91$

Sample of 50 women

This means that $n = 50, s = \frac{2.91}{\sqrt{50}}$

What is the probability that the mean height for the sample is greater than 65 inches?

This is 1 subtracted by the p-value of Z when X = 65. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{65 - 64.4}{\frac{2.91}{\sqrt{50}}}$

$Z = 1.46$

$Z = 1.46$ has a p-value of 0.9279

1 - 0.9279 = 0.0721

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

8 0

2 years ago

Factor this expression: ab + 4 + a + 4b

Makovka662 [10]

Grouping method works best on this one:

ab+a+4+4b=a(b+1)+4(b+1)
=(a+4)(b+1)

4 0

2 years ago

Drum Tight Containers is designing an open-top, square-based, rectangular box that will have a volume of 62.5 in cubed. What di

Orlov [11]

Let the square base of the container be of side s inches and the height of the container be h inches, then
Surface are of the container, A = s^2 + 4sh
For minimum surface area, dA / ds + dA / dh = 0
i.e. 2s + 4h + 4s = 0
6s + 4h = 0
s = -2/3 h

But, volume of container = 62.5 in cubed
i.e. s^2 x h = 62.5
(-2/3 h)^2 x h = 62.5
4/9 h^2 x h = 62.5
4/9 h^3 = 62.5
h^3 = 62.5 x 9/4 = 140.625
h = cube root of (140.625) = 5.2 inches
s = 2/3 h = 3.47

Therefore, the dimensions of the square base of the container is 3.47 inches and the height is 5.2 inches.

The minimum surface area = s^2 + 4sh = (3.47)^2 + 4(3.47)(5.2) = 12.02 + 72.11 = 84.13 square inches.

5 0

2 years ago

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Solve each system of linear equations by substitution ​

Solve each system of linear equations by substitution