X + 5 * 3 divided by 4

6 plates and 5 cups cost 32.10 dollars. 7 plates and 6 cups cost 37.70 dollars. Each plate costs the same amount as the other pl

slavikrds [6]

Answer:

Plates are 2.80 Dollars and cups are 1.40 dollars

Step-by-step explanation:

Cups = c

Plates = p

6p + 5c = 23.80 ==> multiply by 6 ==> 36p + 30c = 142.80

7p + 6c = 28.00 ==> multiply by 5 ==> 35p + 30c = 140.00

Now use algebra to have the 30c be on one side and the rest on the other:

36p + 30c = 142.80 | -36p

30c = 142.80 - 36p

35p + 30c = 140.00 | -35p

30c = 140.00 - 35p

Now set them equal to each other about the 30c:

142.80 - 36p = 140.00 - 35p

Use algebra to solve for p:

142.80 - 36p = 140.00 - 35p | +36p

142.80 = 140.00 + p | - 140

2.80 = p

Go back to one of the "30c" equations and plug in the value for p:

30c = 140 - 35p

30c = 140 - 35(2.80)

30c = 140 - 98

30c = 42 | /30

c = 42/30

c = 1.4

8 0

3 years ago

Read 2 more answers

Traeq pays 35.34 for 3.1 pounds of salmon. What is the price per pound of salmon

Morgarella [4.7K]

$11.40 because you just divide 35.34 by 3.1

3 0

3 years ago

The admission fee to a video game arcade is $1.25 per person, and it costs $0.50 for each game played. Latoya and donnetta

dalvyx [7]

Answer:

15 games.

Step-by-step explanation:

Here is the complete question: The admission fee to a video game arcade is $1.25 per person, and it costs $0.50 for each game played. Latoya and donnetta have total $10.00 to spend. What is the greatest number of games they will be able to play?

Given: Admission fees= $1.25 per person.

Cost of each game played= $0.50.

Total Money, Latoya and Donnetta have is $10.

First lets find total admission fees both will pay.

∴ Total addmission fees= $Fees\ per\ person \times number\ of\ person$

Total addmission fees= $\$1.25\times 2= \$ 2.50$

Now, money remaining after paying admission fees.

Money remaining = $Total\ money - Admission\ fees$

∴ Money remaining = $\$10-\$2.5= \$ 7.5$

∴ Money remained to play video games is $7.5.

Next, finding the maximum number of games, both will be able to play.

As we know, each game will cost $0.50.

Total number of games played= $\frac{Money\ remaining}{Cost\ of\ each\ game}$

⇒ Total number of games played= $\frac{\$ 7.5}{\$ 0.5} = 15\ games$

∴ 15 games is the maximum number of games they will be able to play.

3 0

3 years ago

If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person se

notsponge [240]

Answer:

2.28% probability that a person selected at random will have an IQ of 110 or higher

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 100, \sigma = 5$