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1 year ago

Why does the notation 13c- have a negative sign

1 answer:

6 0

<span>The negative symbol or negative sign simply indicates that for the C ion, there are more electrons than proton. Since it is – alone without number, this automatically indicates a number of -1. This means that there is one more electron than proton.</span>

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Victor fue al mercado para comprar manzanas, naranjas y platanos; las naranjas costaron el doble de lo 1ue pago por las manzanas

Thepotemich [5.8K]

Answer:

El precio de las manzanas = 27 pesos

El precio de las naranjas = 54 pesos

El precio de las bananas = 19 pesos

Step-by-step explanation:

Los parámetros dados son;

El monto total gastado = 100 pesos

Sea el precio de las naranjas = x

Sea el precio de las manzanas = y

Sea el precio de los plátanos = z

La cantidad pagada por las naranjas = 2 · y = x

La cantidad pagada por los plátanos = y - 8 = z

Por lo tanto, tenemos;

La cantidad total gastada = La cantidad pagada por las naranjas + La cantidad pagada por las bananas + La cantidad pagada por las manzanas

∴ El monto total gastado = 100 pesos = 2 · y + y - 8 + y

100 = 4 · años - 8

4 · y = 100 + 8 = 108

y = 108/4 = 27

y = 27

z = y - 8 tenemos;

z = 27 - 8 = 19

De 2 · y = x, tenemos;

2 × 27 = x

x = 54

Por lo tanto;

El precio de las naranjas = 54 pesos

El precio de las manzanas = 27 pesos

El precio de los plátanos = 19 pesos.

5 0

1 year ago

Which equation below shows a correct first step for solving 3x = -12

Wewaii [24]

The answer is A. You just solve each one and you would find that it would be the same as 3x=-12.

5 0

1 year ago

The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin

Free_Kalibri [48]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.03

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03$

$=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12$

Thus, 8.17 g separates the top 3% of the weights.

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90$

Thus, 7.90 separates the bottom 3% of the weights.

Thus, the correct answer is

Option D) 7.90 g and 8.12 g

7 0

2 years ago

The equations below represent the numbers y of tickets sold after x weeks for two different local music festivals

Anton [14]

Assuming you mean y=10x+150 and y=20x+115, you need to use a simultaneous equation, because you have two equations with two unknowns (x and y)

rearrange so
10x-y=-150
20x-y=-115

multiply the top by -1, so that if we add the two lines together, the y will cancel out
-10x+y=150
20x-y=-115

add the two lines together
10x=35
x=3.5

so the time is 3 and a half weeks

then we can sub in x to find y
20x-y=-115
20(3.5)-y=-115
70-y=-115
-y=-185
y=185

so 185 tickets were sold !
you can sub these values into your original equations to check your answer :)

5 0

2 years ago

Read 2 more answers

State, with an explanation, whether the mean, median, or mode gives the best description of the following average. The average

erma4kov [3.2K]

Answer:

The mean is the better method.

Step-by-step explanation:

The best way to meassure the average height is throught mean. The mean of a sample is the average of that sample's height, and it will be a good estimate for the population's average height.

The mode just finds the most frequent height. Even tough the most frequent height will influence the average height, knowing only what height is the most frequent one doesnt give you enough informtation about how the height is centrally distributed.

As for the median, it is fine to use the median of a sample to estimate the median of the population, but if you use the median to estimate the average height you may have a few issues. For example, if you include babies in your population, the babies will push the average height down a lot and they are far below te median height. This, as a result, will give you a median height of a sample way above the average height of the population, becuase median just weights every person's height the same, while average will weight extreme values more, in the sense that a small proportion of extreme values can push the average far from the median.

8 0

1 year ago