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Neko [114]
3 years ago
12

Solve. 7 pounds × 16 ounces = pounds

Mathematics
1 answer:
Anna35 [415]3 years ago
8 0

Answer:

7

Step-by-step explanation:

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Sandra throws an object into the air with an initial vertical velocity of 38 feet per second, from a platform that is 30 feet ab
Blababa [14]
<span>To solve it, use the quadratic formula with (½)(-32.174 ft/s²) = a, 38 ft/s = b, and 30 ft = c. There are two answers; the only positive answer is t = 2.986 s </span>
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3 years ago
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(GIVING BRAINLIYEST)<br> 7. what is the volume of these 2 objects
inessss [21]
3. 140 m^3
(3 x 6 x 7) + (1 x 2 x 7) = 126 + 14 = 140

4. 216 in^3
(4 x 4 x 9) + (3 x 4 x 6) = 144 + 72 = 216
6 0
3 years ago
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What is the answer to <br> 24=3(n−5)
-Dominant- [34]
3(n - 5) = 24

Distributive property

3*n = 3n
3*5 = 15

3n - 15 = 24
      + 15  +15

3n = 39
3n/3 = 39/3
x = 39/3
x =  13

Answer: x = 13
5 0
3 years ago
Which represents where f(x) = g(x)?
son4ous [18]

Answer:

Option (1)

Step-by-step explanation:

Two functions 'f' and 'g' have been graphed in the picture attached.

Both the functions will be equal at the points where the values of these functions are equal.

Those points are the point of intersection of both the functions on the given graph.

At x = -4,

f(-4) = g(-4) = 4

At x = 0,

f(0) = g(0) = 4

Therefore, Option (1) will be the answer.

5 0
3 years ago
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
3 years ago
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