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2 years ago

Please help me Create a system of two equations but do not solve the system please actually help me please and thank you

Mathematics

1 answer:

mel-nik [20]2 years ago

8 0

Answer:

7r + g = 46

7r + 2g = 50

Step-by-step explanation:

Hi there!

Matt and Trevon both bought supplies from a store, and we know that Matt bought 7 rose bushes and 1 geranium, and spent a total of $46 while Trevon bought 7 rose bushes and 2 geraniums, and spent a total of $50

We know that each rose bush costs a certain amount, and each geranium must cost a certain amount, but we don't know what that amount is; it's also possible that the amounts could be different

So let's say that the cost of one rose bush is r dollars; for Matt, that means that he spent 7r dollars on 7 rose bushes. Trevon also bought 7 rose bushes, meaning he also spent 7r dollars on rose bushes.

Since the cost of the geraniums can be different from the cost of every rose bush, let's express it with a different variable, like g.
Since Matt bought one geranium, he spent g dollars on that one geranium.
Trevon bought 2 geraniums, so he spent 2g dollars on the geraniums

The total cost of what Matt spent on flowers was 7r + g dollars (7 rose bushes + 1 geranium), which is equal to 46 dollars

As an equation, that can be written as 7r + g = 46

Trevon spent 7r + 2g dollars (7 rose bushes + 2 geraniums), which gave him a total of $50

As an equation, that can be written as 7r + 2g = 50
The two equations therefore are:
7r + g =46
7r+2g=50
Hope this helps!

Take a look here to practice with a similar problem if you wish (you don't have to look at the actual solving portion): brainly.com/question/14033905

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a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

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