Is anyone able to see if there is a counterexample for the following argument in logic?

Read the problem and answer the questions show your solution when necessary.

Alexandra [31]

Answer:

Step-by-step explanation:

1) Number of white chocolate bars = x

Number of dark chocolate bars = y

Total bars = 15

x + y = 15 --------------(I)

Total cost = Php. 340

20x + 25y = 340 ----------------(II)

2) We can use elimination method to solve this equations.

Multiply equation (I) by (-20) and then add the equations.

(I)* (-20) -20x -20y = - 300

(II) <u>20x + 25y = 340 </u> {Now add and x will be eliminated}

5y = 40

y = 40/5

y = 8

Substitute y = 8 in equation (I)

x + 8 = 15

x = 15 - 8

x = 7

3) Number of dark chocolate bars = 8

4 0

3 years ago

Helppppppppppppppppppppppppp

Advocard [28]

Answer:

$2

$18

Step-by-step explanation:

6 0

3 years ago

The average THC content of marijuana sold on the street is 9.3%. Suppose the THC content is normally distributed with standard d

velikii [3]

Answer:

a) $X \sim N(9.3,1)$

b) $P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

c) The value of height that separates the bottom 75% of data from the top 25% is 9.9745.

Step-by-step explanation:

1) 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".

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(9.3,1)$

Where $\mu=9.3$ and $\sigma=1$

3) Part b

We are interested on this probability

$P(X>9.2)$

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

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

If we apply this formula to our probability we got this:

$P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

4) Part c

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.6745. On this case P(Z<0.6745)=0.75 and P(z>0.6745)=0.25

If we use condition (b) from previous we have this:

$P(X$