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Vadim26 [7]
2 years ago
15

In creating the shooter-style game comparable to Space Invaders, we decided that each time the laser hits an object, the player

gets a point. But if the object falls on the ship, the player loses a life.Which element of a game are we describing?
Group of answer choices

operation

objective

obstacle

outcome
Mathematics
1 answer:
olya-2409 [2.1K]2 years ago
3 0

Answer:

Outcome

Step-by-step explanation:

I'm not exactly certain but I believe the answer is outcome. For example: Each time the laser hits an object, the player gets a point. Because you're giving the cause, the laser hitting the object, and the effect, the player getting a point;its safe to assume that the element is outcome.

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In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
In Vancouver, British Columbia, the probability of rain during a winter day is 0.42, for a spring day is 0.23, for a summer day
nadezda [96]

Answer:

31.82% probability that this day would be a winter day

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening

In this question:

Event A: Rain

Event B: Winter day

Probability of rain:

0.42 of 0.25(winter), 0.23 of 0.25(spring), 0.16 of 0.25(summer) or 0.51 of 0.25(fall).

So

P(A) = 0.42*0.25 + 0.23*0.25 + 0.16*0.25 + 0.51*0.25 = 0.33

Intersection:

Rain on a winter day, which is 0.42 of 0.25. So

P(A \cap B) = 0.42*0.25 = 0.105

If you were told that on a particular day it was raining in Vancouver, what would be the probability that this day would be a winter day?

P(B|A) = \frac{0.105}{0.33} = 0.3182

31.82% probability that this day would be a winter day

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Answer:

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Step-by-step explanation:


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Answer:

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Step-by-step explanation:

In the table, r = 3p, and by converse, p = 1/3r. These options are D and E

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Step-by-step explanation:

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The result is CAD 708.76

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