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Oksana_A [137]
3 years ago
13

You want to buy a tent in the shape of a pyramid. The rectangular base is 35 square feet, with a height of 4 feet. What is the v

olume of the tent? Round your answer to the nearest hundredth.
Mathematics
1 answer:
marin [14]3 years ago
7 0

Answer:

46.67 cubic feet

Step-by-step explanation:

The formula for volume of a pyramid is

V = (1/3)Bh      where B is the area of the base, and h is the height.

We are givne B = 35 and h = 4.  Plug those values in and simplify...

V = (1/3)(35)(4)

 

  V = (1/3)(140)

       V = 140/3

             V = 46 2/3 cubic feet, which is 46.666667, or 46.67 rounded to the nearest hundredth

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The answer i got was
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7 0
3 years ago
QUESTION
Lorico [155]
1. For this item we just refer to the prompt to know the conjectures of Ernest and Denise. According to Ernest, they should swim 1 kilometer on the first week then add 0.25km every week while Denise believes that they should swim 1 kilometer on the first week then add 0.5km every week.

2. Yes, these distances make an arithmetic sequence. It's because an arithmetic sequence is defined as a group of increasing or decreasing numbers where the difference between any two consecutive numbers is constant. This just means that every number has the same interval. In the case of their schedule, this is true.

3. For this item we just follow the descriptions of Ernest's and Denise's schedule in item number 1. For Ernest, we just keep adding 0.25 from 1 kilometer until we added it thrice. For Denise, we also keep adding a number thrice but this time it's 0.5 instead of 0.25.

Ernest's Schedule: 1, 1.25, 1.5, 1.75
Denise's Schedule: 1, 1.5, 2, 2.5

4. Here we are asked to determine a formula that will describe the schedules of Ernest and Denise. In the given formula a_{n}= a_{n-1}+d, a_{n} refers to the next term in the sequence, a_{n-1} refers to the previous term, while d refers to the common difference. In the recursive formula all we need is to insert the value of d to the equations.

Ernest: a_{n}= a_{n-1}+0.25
Denise: a_{n}= a_{n-1}+0.5

5. For this item we basically do the same thing but this time we are given another formula. Our formula is in the form a_{n}= a_{1}+(n-1)d where a_{n} is still the nth term of the sequence, a_{1} is the very first time, n is the number of terms, and d is the common difference. 

Ernest: a_{n}= 1.0+0.25(n-1)
Denise: a_{n}= 1.0+0.5(n-1)

6. In this item we will just basically substitute numbers to one of the equations that we've set up in item #5. For this we need Ernest's explicit formula first. To know how far they will be swimming on week 10, the number of elements (n) must be 10.

a_{10}= 1.0+0.25(10-1)
a_{10}= 1.0+0.25(9)
a_{10}= 1.0+2.25
a_{10}= 3.25

7. Here, we just do the same thing as item #6 but this time we will consider Denise's explicit formula. Since we are also asked how far the students will be swimming on week 10, the number of elements would also be 10 and this would also be our value for n.

a_{10}= 1.0+0.5(10-1)
a_{10}= 1.0+0.5(9)
a_{10}= 1.0+4.5
a_{10}= 5.5

8. The answer for this question is obvious. You would just need to look at the 10th element in Ernest's and Denise's sequences and tell whose schedule had more than or equal to 5 as an answer. Following Ernest's schedule, you will just get 3.5 kilometers on the 10th week so it's definitely a no. Denise's schedule, on the other hand, would get you to 5.5 kilometers on week 10 so her training schedule should be followed.
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Answer:

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Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round yo
ycow [4]

Answer:

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

n = 201, p = 0.45

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

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Answer:

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Step-by-step explanation:

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