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vredina [299]
2 years ago
5

When filled, a balloon in the shape of a sphere has a diameter of 14 inches. What is the approximate volume of air, in cubic inc

hes, needed to fill this balloon?
1,078
1,437
8,621
11,494 ​
Mathematics
2 answers:
UkoKoshka [18]2 years ago
3 0

Answer:

1,437

Step-by-step explanation:

ziro4ka [17]2 years ago
3 0

Answer:

a I'm not as smart so trust me

Step-by-step explanation:

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Evaluate the expression 2ab 3+ 6a 3 - 4ab2 when a=2 and b=4.<br> a.176<br> b.180<br> c.215<br> d.216
Tom [10]
The first step for solving this expression is to insert what a and b stand for into the expression. This will change the expression to the following:
2(2)(4)³ + 6(2)³ - 4(2)(4)²
Now we can start solving this by factoring the expression
2(2 × 4³ + 3 × 2³ - 4 × 4²)
Write 4³ in exponential form with a base of 2.
2(2 × 2^{6} + 3 × 2³ - 4 × 4²)
Calculate the product of -4 × 4².
2(2 × 2^{6} + 3 × 2³ -4³)
Now write 4³ in exponential form with a base of 2.
2(2 × 2^{6} + 3 × 2³ -2^{6})
Collect the like terms with a base of 2.
2(2^{6} + 3 × 2³)
Evaluate the power of 2³.
2(2^{6} + 3 × 8)
Evaluate the power of 2^{6}.
2(64 + 3 × 8)
Multiply the numbers.
2(64 + 24)
Add the numbers in the parenthesis.
2 × 88
Multiply the numbers together to find your final answer.
176
This means that the correct answer to your question is option A.
Let me know if you have any further questions.
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5 0
3 years ago
Plzzz help!!!!!!<br> Write y=-3/5x -2 in standard form using integers.
Delicious77 [7]
That is the answer , i looked it up for you my self :)

7 0
3 years ago
Read 2 more answers
The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

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7 0
3 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
A student found out that 8% ofthe students in his school are left-handed.If there are 114 lefties, how many students are in his
AleksAgata [21]

Answer: 1,425 students

Step-by-step explanation: This is easy to solve using proportions/

114/8 = ?/100

First, you cross multiply (or whatever this is called sry) 100x114=11400

Then divide 11400÷8=1,425

8 0
2 years ago
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