We'll need one equation for the amount of pizzas, and another for the total cost of the pizzas. In this case, x will represent small pizzas, and y will represent large pizzas.

3x + 4y = 100

x + y = 30

Solving the system of equations:

First, we need to solve for one variable in one equation.

x + y = 30

x = 30 - y

Then, we'll take our equation that is solved for x and plug it into the other equation from above.

3(30 - y) + 4y = 100

Next, we solve for y.

90 - 3y + 4y = 100

90 + y = 100

y = 10

Finally, we take our value for y and plug it back in to the very first equation and solve for x.

x = 30 - 10

x = 20

Answer:

The student has sold 10 large pizzas and 20 small pizzas.

Hope this helps!! :)

6 0

3 years ago

Thank you for your help>>>>>>>>>>+++++++

Shkiper50 [21]

Answer:

Step-by-step explanation:

5 0

3 years ago

Na atmosfera,a temperatura diminui cerca de1 grau a cada 200 m de afastamento da superficie terrestre.Se a temperatura na superf

Alenkinab [10]

Responder:

- 30 ° C

Explicação passo a passo:

Dado que :

A temperatura diminui 1 ° para cada 200 m afastados da superfície terrestre

Temperatura na superfície da terra = 20 ° C

Temperatura atmosférica a uma altura de 10 km de. Superfície Terrestre :

Queda de temperatura para uma altura de 10 km de distância;

10km = 10 * 1000 = 10.000 m

1 * (10000/200)

1 * 50

= 50 ° C

Portanto ;

Temperatura da superfície terrestre - 50 ° C

20 ° C - 50 ° C

= - 30 ° C

4 0

3 years ago

Refer to the Teaching Psychology (May 1998) study of how external clues influence performance. Two different forms of a midterm

kozerog [31]

Answer:

$P(X$

$P(Y$

So for this case we have a higher probability for the red exam so we can conclude that the student is more likely to score below 20% on the difficult questions in the red one.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores for the blue exam, and for this case we know the distribution for X is given by:

$X \sim N(53,15)$

Where $\mu=53$ and $\sigma=15$

We are interested in the probability that P(X<20%)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

Let Y the random variable that represent the scores for the red exam, and for this case we know the distribution for X is given by:

$Y \sim N(39,12)$

Where $\mu=39$ and $\sigma=12$

We are interested in the probability that P(Y<20%)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{y-\mu}{\sigma}$

And if we replace we got:

$P(Y$

So for this case we have a higher probability for the red exam so we can conclude that the student is more likely to score below 20% on the difficult questions in the red one.

7 0

4 years ago