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3 years ago

This image shows a square pyramid.

Mathematics

2 answers:

AURORKA [14]3 years ago

5 0

Answer:

128

Step-by-step explanation:

I just took the test and I can confirm that this answer is correct! :D

Lynna [10]3 years ago

3 0

We know that

[Surface area ]=[the area of 4 triangles]+[the area of square base]
[the area of square base]=8*8--------> 64 ft²

[the area of 1 triangle]=(1/2)*b*h
b=8 ft
How the angle is 45°-------> h=b/2-------> h=4 ft
then
[the area of 1 triangle]=(1/2)*8*4------> 16 ft²
and
[the area of 4 triangles]=4*16--------> 64 ft²

[Surface area ]=[64 ft²]+[64 ft²]----------> 128 ft²

the answer is 128 ft²

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n an orchid show, seven orchids are to be placed along one side of the greenhouse. of the orchids, we see that the number of lin

NNADVOKAT [17]

Answer:

Step-by-step explanation:

7 orchids can be lined as 7!. This means that for the first orchid of the line, you can select 7 options. When you place the first orchid, for the second option you can select among 6 since 1 orchid has already been placed. Similarly, for the 3rd orchid of the line, you have left 5 options. The sequence goes in this fashion and for 7 orchids, you have 7*6*5*4*3*2*1 possibilities. However, there is a restriction here. 3 of the orchids are white and 4 are levender. This means that it does not make a difference if we line 3 white orchids in an arbitrary order since it will seem the same from the outside. As a result, the options for lining the 7 orchids diminish. The reduction should eliminate the number of different lining within the same colors. Similar to 7! explanation above, 3 white orchids can be lined as 3! and 4 levender orchids can be lined as 4!. To eliminate these options, we divide all options by the restrictions. The result is:

$\frac{7!}{4!*3!}$ = 35. [(7*6*5*4*3*2*1/(4*3*2*1*3*2*1)]

3 0

3 years ago

Please Solve This Problem For Me x^2 + 5^2 = 7^2 (^2 Means Squared)

Triss [41]

X^2+5^2=7^2
x^2+25=49
(49-25)
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sqrt(24)=4.89897948557

7 0

3 years ago

Multiple-choice questions on Advanced Placement exams have five options: A, B, C, D, and E. A random sample of the correct choic

Semenov [28]

Answer:

Option B is not the most common correct choice.

Step-by-step explanation:

The multiple-choice questions on Advanced Placement exams have five options: A, B, C, D, and E.

The probability that any of these five option is the correct answer is:

$p=\frac{1}{5}=0.20$

A random sample of 400 multiple-choice questions on Advanced Placement exam are selected.

The results showed that 90 of the 400 questions having B as the answer.

To test the hypothesis that option B is more likely the correct answer for most question, the hypothesis can be defined as:

H₀: All the options are equally probable, i.e. p = 0.20.

Hₐ: Option B is more likely the correct option, i.e. p > 0.20.

Compute the sample proportion as follows:

$\hat p=\frac{90}{400}=0.225$

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.225-0.20}{\sqrt{\frac{0.20(1-0.20)}{400}}}= 1.25$

The test statistic is 1.25.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

$p-value=P(Z>1.25)\\=1-P(Z$

*Use a z-table.

The p-value is 0.1057.

The p-value of the test is quite large. Thus, the null hypothesis was failed to rejected.

Hence, it can be concluded that option B is not the most common correct choice.

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2 years ago

-2(-5x+5) - 3x + 4= -34 solve for x

geniusboy [140]

Answer:

x = -4

Step-by-step explanation:

-2(-5x+5) - 3x + 4= -34

10x - 10 -3x + 4 = -34

7x - 6 = -34

+6 + 6

7x = -28

x = -4

4 0

2 years ago

Read 2 more answers

Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho

oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

7 0

3 years ago