Tobeg says the equation 3x + 2y= 270 and y=4/5x + 20 intersect at the point (50,60) is this true how do you know

Mathematics

1 answer:

il63 [147K]4 years ago

3 0

ANSWER

Yes it is very true

EXPLANATION

If the two equations intersect at $(50,60)$ then this point must satisfy the two equations.

$3x+2y=270---(1)$

We substitute $(50,60)$ in to erquation (1)

$3(50)+2(60)=270$

$150+120=270$

$270=270$

$y=\frac{4}{5}x+20 ---(2)$

We now substitute $(50,60)$ in to erquation (2) also

$60=\frac{4}{5}(50)+20$

$60=40+20$

$60=60$

Since the point satisfy all the two equations, it is true that they intersect at $(60,50)$

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The mean number of hours per day spent on the phone, according to a national survey, is four hours, with a standard deviation of

Reika [66]

Answer:

The new mean is 5.

The new standard deviation is also 2.

Step-by-step explanation:

Let the sample space of hours be as follows, S = {x₁, x₂, x₃...xₙ}

The mean of this sample is 4. That is, $\bar x=\frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}=4$

The standard deviation of this sample is 2. That is, $s=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}=2$ .

Now it is stated that each of the sample values was increased by 1 hour.

The new sample is: S = {x₁ + 1, x₂ + 1, x₃ + 1...xₙ + 1}

Compute the mean of this sample as follows:

$\bar x_{N}=\frac{x_{1}+1+x_{2}+1+x_{3}+1+...+x_{n}+1}{n}\\=\frac{(x_{1}+x_{2}+x_{3}+...+x_{n})}{n}+\frac{(1+1+1+...n\ times)}{n}\\=\bar x+1\\=4+1\\=5$

The new mean is 5.

Compute the standard deviation of this sample as follows:

$s_{N}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{n-1}\sum ((x_{i}+1)-(\bar x+1))^{2}\\=\frac{1}{n-1}\sum (x_{i}+1-\bar x-1)^{2}\\=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=s$