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labwork [276]
3 years ago
7

If (x-2)(y-1)=3 and (x+2)(2y-5)=15,

Mathematics
1 answer:
daser333 [38]3 years ago
5 0

Answer:

Equation 1, Multiplied by 2: 2x + 4y = 10

Equation 1, Multiplied by 2: 2x + 4y - 10 = 10 - 10

Equation 1, Multiplied by 2: 2x + 4y - 10 = 0

Equation 1, Multiplied by 2: 2x - 2x + 4y - 10 = 0 - 2x

Equation 1, Multiplied by 2: 4y - 10 = - 2x

Equation 1, Multiplied by 2: - 10 + 4y = - 2x

Notice that Equation 1 = Equation 2

One Solution

Equation 1: x - y = -10

Equation 2: 2x + 3y = 15

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In a survey, 28 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
Rina8888 [55]

Answer:

The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution should be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 28 - 1 = 27

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.8}{2} = 0.9. So we have T = 1.314

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.314\frac{2.6}{\sqrt{28}} = 0.646

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 34.6 - 0.646 = $33.954

The upper end of the interval is the sample mean added to M. So it is 34.6 + 0.646 = $35.246

The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.

3 0
3 years ago
Dok = (6, 12) (3,6)<br> The scale factor is<br> A. 1/2<br> B. 2<br> C. 4
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Answer:

A. 1/2

Step-by-step explanation:

For 6 to become 3 and 12 to become 6, the (6,12) is being multiplied by 1/2.

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We will use the distributive property, where a(b+c) = ab+ac and a(b-c) = ab-ac.

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