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3 years ago

Do these equations have infinite solutions 2x + 5y= 31 and 6x -y = 13

Mathematics

1 answer:

katovenus [111]3 years ago

8 0

$2x = 31 - 5y \\ x = \frac{31}{2} - \frac{5}{2}y$

$6x = 13 + y \\ x = \frac{13}{6} + \frac{1}{6} y$

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your local gas stations are having a price war during the past seven days they have lowered the price of regular gas by $0.2 eac

Serga [27]

<h2>Answer with explanation:</h2>

Let x denotes the number of days and y denotes the total change in gas price ( in dollars).

Given : During the past seven days the price of regular gas is decreased by $0.2 each day.

Rate of decrease =$0.2 per day

Then , Total change in gas price = Rate of decrease x Number of days

i.e. $y= 0.2 \times x$

i.e. $y=0.2x$ (1)

To find the total change in gas price in 7 days , we put x= 7 in (1) , we get

$y=0.2(7)=1.4$

Hence, the total change in gas price in 7 days = $1.4

8 0

3 years ago

There are 40 GT students going on a field trip. Each one pays their teacher $8.75 to cover admission to the museum and lunch. Ad

Norma-Jean [14]

Step-by-step explanation:

There are 40 DuBois students going on a field trip.

Each one pays their teacher $8.75 to cover admission to the museum and lunch.

Admission for the students is $120 and each one gets an equal amount to spend on lunch.

To Find :

How much will each DuBois student be able to spend on lunch.

Solution :

Amount they paid for admission :

\begin{gathered}P =\$ \dfrac{120}{40}\\\\P=\$3\end{gathered}P=$40120P=$3

Now, amount of money to spend on lunch is 8.75 -8.75−3 = $5.75 .

Hence, this is the required solution.

3 0

3 years ago

PLZ HELP….

MrRissso [65]

The answer should be 1003.53

6 0

3 years ago

Read 2 more answers

Determine the probability of selecting a two-child family with one boy and one girl assuming boys and girls are equally likely

Fittoniya [83]

Using the binomial distribution, it is found that there is a 0.5 = 50% probability of selecting a two-child family with one boy and one girl.

For each child, there are only two possible outcomes, either it is a boy, or it is a girl. The probability of a child being a boy or being a girl is independent of any other child, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

Two children, hence $n = 2$ .
Equally as likely to be a boy or a girl, hence $p = 0.5$ .

The probability of one of each is P(X = 1), hence:

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

$P(X = 1) = C_{2,1}.(0.5)^{1}.(0.5)^{1} = 0.5$

0.5 = 50% probability of selecting a two-child family with one boy and one girl.

A similar problem is given at brainly.com/question/24863377

3 0

3 years ago

Century Go...

LekaFEV [45]

6 bills in the deposit were $10 bills.

Step-by-step explanation:

The total number of bills he gave the teller = 24 bills

x + y = 24 ------------(1)

Let,

x be the number of $5 bills.

y be the number of $10 bills.

The total bank deposit made using $5 and $10 bills are $150.

He deposits $5 bills and $10 bills. So, the equation can be framed as :

5x + 10y = 150 --------(2)

The question is asked for how many $10 bills were in the deposit,

Here 'y' is the number of $10 bills.

To solve for y value :

The eq(1) must be multiplied by 5 and subtract eq(2) from eq(1),

5x + 5y = 120

-(5x + 10y = 150)

-5y = -30

⇒ y = 30/5

⇒ y = 6

Therefore, there were 6 bills in the deposit that are $10 bills.

3 0

3 years ago