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2 years ago

If 2x – 10y = -1 and 5x + 6y = 4 are true equations, what would be the value of 7x - 4y?

Mathematics

1 answer:

Zigmanuir [339]2 years ago

8 0

Answer:

Step-by-step explanation:

If the equations are true, they can be solved simultaneously.

Consider 2x-10y=-1 as Eq1 and 5x+6y=4 as Eq2

Multiplying Eq1 with 3 and Eq2 with 5 we get,

6x-30y=-3 -- Eq3 and 25x+30y=20 -- Eq4

Adding Eq3 and Eq4,

31x=17 Therefore, x=17/31

Plugin x=17/31 in Eq3,

6(17/31)-30y=-3 --- y=13/62

Plug In values of x and y in 7x-4y we get,

7(17/31)-4(13/62) = 3

So the value of 7x-4y = 3.

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3 years ago

Points D and E are midpoints of the sides of triangle ABC. The perimeter of the triangle is 48 units. Triangle A B C is shown. P

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Answer:

Step-by-step explanation:

From the diagram attached :

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3 years ago

Read 2 more answers

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Taya2010 [7]

Answer:

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3 years ago

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. His

morpeh [17]

Answer:

There is a 21.053% probability that this person made a day visit.

There is a 39.474% probability that this person made a one night visit.

There is a 39.474% probability that this person made a two night visit.

Step-by-step explanation:

We have these following percentages

20% select a day visit

50% select a one-night visit

30% select a two-night visit

40% of the day visitors make a purchase

30% of one night visitors make a purchase

50% of two night visitors make a purchase

The first step to solve this problem is finding the probability that a randomly selected visitor makes a purchase. So:

$P = 0.2(0.4) + 0.5(0.3) + 0.3(0.5) = 0.38$

There is a 38% probability that a randomly selected visitor makes a purchase.

Now, as for the questions, we can formulate them as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

Suppose a visitor is randomly selected and is found to have made a purchase.

How likely is it that this person made a day visit?

What is the probability that this person made a day visit, given that she made a purchase?

P(B) is the probability that the person made a day visit. So $P(B) = 0.20$

P(A/B) is the probability that the person who made a day visit made a purchase. So $P(A/B) = 0.4$

P(A) is the probability that the person made a purchase. So $P(A) = 0.38$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.4*0.2}{0.38} = 0.21053$

There is a 21.053% probability that this person made a day visit.

How likely is it that this person made a one-night visit?

What is the probability that this person made a one night visit, given that she made a purchase?

P(B) is the probability that the person made a one night visit. So $P(B) = 0.50$

P(A/B) is the probability that the person who made a one night visit made a purchase. So $P(A/B) = 0.3$

P(A) is the probability that the person made a purchase. So $P(A) = 0.38$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.3}{0.38} = 0.39474$

There is a 39.474% probability that this person made a one night visit.

How likely is it that this person made a two-night visit?

What is the probability that this person made a two night visit, given that she made a purchase?

P(B) is the probability that the person made a two night visit. So $P(B) = 0.30$

P(A/B) is the probability that the person who made a two night visit made a purchase. So $P(A/B) = 0.5$

P(A) is the probability that the person made a purchase. So $P(A) = 0.38$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.5}{0.38} = 0.39474$

There is a 39.474% probability that this person made a two night visit.

3 0

4 years ago