28. Surface Area
This is some sort of house-like model so for every face we see there's a congruent one that's hidden. We'll just double the area we can see.
Area = 2 × ( [14×9 rectangle] + 2[15×9 rectangle]+[triangle base 14, height 6] )
Let's separate the area into the area of the front and the sides; the front will help us for problem 29.
Front = [14×9 rectangle] + [triangle base 14, height 6]
= 14×9 + (1/2)(14)(6) = 14(9 + 3) = 14×12 = 168 sq ft
OneSide = 2[15×9 rectangle] = 30×9 = 270 sq ft
Surface Area = 2(168 + 270) = 876 sq ft
Answer: D) 876 sq ft
29. Volume of an extruded shape is area of the base, here the front, times the height, here 15 feet.
Volume = 168 * 15 = 2520 cubic ft
Answer: D) 2520 cubic ft
4x+4(3)+4(2)=y
4x+20=y
X equals the amount it cost for each ticket which is unknown. Y equals the total cost. Because the amount of the food is provided, it can be calculated unlike the tickets which is just left at 4x. The work for the snacks are upwards. Its mostly substitution.
Based on the information in the table, an example of independent events is the: P(policer officer and chooses car).
<h3>What is an
independent event?</h3>
An independent event can be defined as an event that isn't dependent on other events. Thus, it isn't affected by any previous event.
Based on the information provided, we can infer and logically deduce that an example of independent events is the probability of being a policer officer and chooses a car because they aren't overlapping probabilities.
Read more on independent events here: brainly.com/question/26795996
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Answer:
H0: The distribution of players featured on the cards is 0.30 rookies, 0.60 veterans, and 0.10 All-Stars.
Ha: At least one of the proportions in the null hypothesis is false.
Step-by-step explanation:
On this case we need to apply a Chi squared goodness of fit test, and the correct system of hypothesis would be:
H0: The distribution of players featured on the cards is 0.30 rookies, 0.60 veterans, and 0.10 All-Stars.
Ha: At least one of the proportions in the null hypothesis is false.
And in order to test it we need to have observed and expected values. On this case we can calculate the Expected values like this
The observed values are not provided. The statistic on this case is given by:
And this statistic follows a chi square distribution with k-1 degrees of freedom on this case k=3, since we have 3 groups.
We can calculate the p valu like this:
And if the p value it's higher than the significance level we FAIL to reject the null hypothesis. In other case we reject the null hypothesis.
(12 x 15)m< 68
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I hope this helps! I couldn’t get the line below the sign lol. Sorry if wrong
