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schepotkina [342]

3 years ago

11

A flying fish deep under the sea dreams to reach the clouds one day the fish is at a depth of 200 feet below sea level right und

er the clouds it has to fly 6700 feet to reach the clouds how many feet above sea level are the clouds

1 answer:

Anarel [89]3 years ago

6 0

Answer:

6500

Step-by-step explanation:

6700-200= 6500

6500 ft above sea level

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A pyramid has a square base of length 8cm and a total surface area of 144cm². Find the volume of the pyramid. (Please use Pythag

Sveta_85 [38]

Answer:

$\displaystyle V_{ \text{pyramid}}= 64 \: {cm}^{3}$

Step-by-step explanation:

we are given surface area and the length of the square base

we want to figure out the Volume

to do so

we need to figure out slant length first

recall the formula of surface area

$\displaystyle A_{\text{surface}}=B+\dfrac{1}{2}\times P \times s$

where B stands for Base area

and P for Base Parimeter

so

$\sf\displaystyle \: 144=(8 \times 8)+\dfrac{1}{2}\times (8 \times 4) \times s$

now we need our algebraic skills to figure out s

simplify parentheses:

$\sf\displaystyle \: 64+\dfrac{1}{2}\times32\times s = 144$

reduce fraction:

$\sf\displaystyle \: 64+\dfrac{1}{ \cancel{ \: 2}}\times \cancel{32} \: ^{16} \times s = 144 \\ 64 + 16 \times s = 144$

simplify multiplication:

$\displaystyle \: 16s + 64 = 144$

cancel 64 from both sides;

$\displaystyle \: 16s = 80$

divide both sides by 16:

$\displaystyle \: \therefore \: s = 5$

now we'll use Pythagoras theorem to figure out height

according to the theorem

$\displaystyle \: {h}^{2} + (\frac{l}{2} {)}^{2} = {s}^{2}$

substitute the value of l and s:

$\displaystyle \: {h}^{2} + (\frac{8}{2} {)}^{2} = {5}^{2}$

simplify parentheses:

$\displaystyle \: {h}^{2} + (4 {)}^{2} = {5}^{2}$

simplify squares:

$\displaystyle \: {h}^{2} + 16 = 25$

cancel 16 from both sides:

$\displaystyle \: {h}^{2} = 9$

square root both sides:

$\displaystyle \: \therefore \: {h}^{} = 3$

recall the formula of a square pyramid

$\displaystyle V_{pyramid}=\dfrac{1}{3}\times A\times h$

where A stands for Base area (l²)

substitute the value of h and l:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{3}\times \{8 \times 8 \}\times 3$

simplify multiplication:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{3}\times 64\times 3$

reduce fraction:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{ \cancel{ 3 \: }}\times 64\times \cancel{ \: 3}$

hence,

$\sf\displaystyle V_{ \text{pyramid}}= 64 \: {cm}^{3}$

8 0

3 years ago

What is the answer? <br><br><br> A: 5/20<br><br> B:5/11<br><br> C: 19/20<br><br> D: 1 9/20

Kaylis [27]

The answer is D.
Find common factors of denominators.
15+10+4/20= 29/20= 1 9/20 :)

7 0

3 years ago

The probability of a customer arrival at a grocery service counter in any one second is equal to 0.3. Assume that customers arri

Svetllana [295]

Answer:

Step-by-step explanation:

Given that the probability of a customer arrival at a grocery service counter in any one second is equal to 0.3

Assume that customers arrive in a random stream, so an arrival in any one second is independent of all others.

i.e. X the no of customers arriving is binomial with p = 0.3 and q = 1-0.3 =0.7

a) the probability that the first arrival will occur during the third one-second interval.

= Prob that customer did not arrive in first 2 seconds * prob customer arrive in 3rd sec

= $0.7^2 (0.3)\\= 0.147$

b) the probability that the first arrival will not occur until at least the third one-second interval.

Prob that customer did not arrive in first two seconds *(Prob customer arrives in 3rd or 4th or 5th.....)

= $(0.7^2)[0.3+0.7*0.3+0.7^2*0.3+....)\\$

The term inside bracket is a geometric infinite progression with common ratio - 0.7 <1

Hence the series converges

Prob = $0.7^2 *\frac{0.3}{1-0.7} \\=0.49$

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3 years ago

I need help with linear functions can anybody help me?

NNADVOKAT [17]

I can help you! Just let me how what the questions are

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A. A number line with an open circle on -4, shading to the left, and a closed circle on 3, shading to the right

8 0

3 years ago

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