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3 years ago

What is the solution to the given system of equations?

Mathematics

2 answers:

marin [14]3 years ago

8 0

Hi!

See my image for the answer.

I put each of the choices into the equation, then solved it.

The answer is B. (4,-12)

Hope this helps! :)
-Peredhel

lukranit [14]3 years ago

6 0

If both those equations are equal to y, then that means by the transitive property, they are equal to each other. We set them equal to each other and solve for x. 4x - 28 = 1.5x - 18. Subtract 1.5x from both sides to get 2.5x = 10 and x = 4. Now sub that value in for x into one of the original equations to find y: y=4(4)-28 and y = 16 - 28 so y = -12. Your solution to that system is (4, -12), B

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Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd =

swat32

Answer:

Test statistic, $t_{s} = -0.603$ (to 3 dp)

Step-by-step explanation:

Deviation, d = x -y

Sample mean for the deviation

$\bar{d} = \frac{\sum x-y}{n}$

$\bar{d} = \frac{(28-6) + (31-27)+(20-26)+(25-25)+(28-29)+(27-32)+(33-33)+(35-34)}{8} \\\bar{d} = -0.625$

Standard deviation: $SD = \sqrt{\frac{\sum d^{2} - n \bar{d}^2}{n-1} }$

$\sum d^{2} = (28-26)^2 + (31-27)^2 +(20-26)^2 +(25-25)^2 +(28-29)^2 +(27-32)^2 +(33-33)^2 +(35-34)^2\\\sum d^{2} = 63$

$SD = \sqrt{\frac{63 - 8 * (-0.625)^2}{8-1} }$

SD =2.93

Under the null hypothesis, the formula for the test statistics will be given by:

$t_{s} = \frac{ \bar{d}}{s_{d}/\sqrt{n} } \\t_{s} = \frac{- 0.625}{2.93/\sqrt{8} }$

$t_{s} = -0.6033$

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3 years ago

Concrete can be purchased by the cubic yard. How much will it cost to pour a slab 18 feet by 18 feet by 6 inches for a patio if

lesya692 [45]

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3 years ago

Read 2 more answers

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.