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mamaluj [8]
3 years ago
15

In a video game, the player can choose a character from 4 animal characters and 3 human characters. Players can also let the com

puter randomly select a character for them.
The player will have the computer randomly select his character.
What is the probability it will pick a human character?
Enter your answer as a fraction in simplest form in the box.
Mathematics
2 answers:
Elis [28]3 years ago
6 0

Answer:

The answer is 3/7 (Three over seven). You ADD the number of total possibilities (animal characters + human characters = 7 total characters) for the denominator and put in the number (or section) you're looking at as the numerator

Hope this helps! :)

Zepler [3.9K]3 years ago
4 0

Answer: 3/7

Step-by-step explanation: Adding 4 + 3 equals 7 so the probability of the computer picking a human character over a animal character is 3/7.

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3 years ago
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Assoli18 [71]

Answer:

% error = 3.3%

Step-by-step explanation:

Percent Error Formula:

% error = ( (MeasuredValue - AcceptedValue) / AcceptedValue ) * 100

% error = ( (12.7 - 12.3) / 12.3 ) * 100

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Free_Kalibri [48]

k(-3)=36(2)

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7 0
2 years ago
AC if TC = 20q + 10q^2?
Alexus [3.1K]

Answer:

AC = (20+ 10q)

Step-by-step explanation:

Given that,

Total cost, TC = 20q + 10q²

We need to find AC i.e. average cost.

It can be solved as follows :

AC=\dfrac{TC}{q}\\\\AC=\dfrac{20q + 10q^2}{q}\\\\AC=\dfrac{q(20+ 10q)}{q}\\\\AC={(20+ 10q)}

So, the value of AC is (20+ 10q).

4 0
3 years ago
According to the University of Nevada Center for Logistics Management, 6% of all mer-
Fudgin [204]

Answer:

a) The point estimate of the proportion of items returned for the population of

sales transactions at the Houston store = 12/80 = 0.15

b) The 95% confidence interval for the proportion of returns at the Houston store = [0.0718 < p < 0.2282].

c) Yes.

We set an hypothesis and construct a test statistics. The test statistics result gives us:

Z calculated  = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

Step-by-step explanation:

a) Point estimate of the proportion = number of returned items/ total items sold = 12/80 = 0.15.

b) By formula of confident interval:

CI(95%) = p ± Z*\sqrt{\frac{p*(1-p)}{n} }  =  0.15 \pm 1.96 *\sqrt{\frac{0.15*(1-0.15)}{80} },

CI(95%) = [0.0718 < p < 0.2282]

c) The hypothesis:

H_{0}: The proportion of returns at the Houston store is not significantly different from the returns  for the nation as a whole.

H_{a}: The proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

The test statistics:

Z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p*(1-p)}{n} }}, where p_{0} is the proportion of nation returns.

Z calculated  = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

6 0
3 years ago
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