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Leokris [45]
3 years ago
14

Dale is in a ski rental shop trying to decide which equipment to rent for the day. The shop offers 4 kinds of skis and 3 kinds o

f poles. A helmet is always a good idea, and the shop has 8 different helmets available. How many different sets of ski equipment can Dale rent? sets
Mathematics
2 answers:
11111nata11111 [884]3 years ago
8 0

Answer: 12 poles and 16 helmets

Step-by-step explanation:

3+4 =8

-4= -4

1.7

Hoochie [10]3 years ago
7 0

Answer:

96

Step-by-step explanation:

Simply multiply all the numbers together.

4 * 3 = 12

12 * 8 = 96

There are 96 possible combinations.

If this helped, please mark brainliest :D

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Whos the best airbender ever lol
Sholpan [36]

Answer:

me

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The table shows the masses of four objects. A 2-column table with 4 rows. The first column labeled object has entries book, rock
Ede4ka [16]

Answer:

ROCK 2.3 kg

is about 22.5N

Step-by-step explanation:

To convert kg>>>>N

You multiply by 9.8 (this is good for estimating, but use more decimal places like, 9.80665 for more exact calculations).

You can divide N÷9.8 to find kg

see image

6 0
2 years ago
Josh is solving the equation 2 % + 2% and predicts that his
UkoKoshka [18]

Answer: 24 is the LCD

He is correct.

4 20/24 or 4 5/6 simplified

4 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
Which expressions are equivalent to the expression below?
Dmitriy789 [7]

Given :-

  • -1/3 - ( -4 + 1/6 )

To Find :-

  • The expression that is equivalent to one of the choices given .

Solution :-

As we know that ,

→ (-) × (-) = (+)

→ (-) × (+) = (-)

→ (+) × (-) = (-)

→ (+) × (+) = (+)

On using these open the brackets ,

→ -1/3 - 1( -4 + 1/6 )

→ -1/3 - 1(-4) + -1(+1/6)

→ -1/3 (-)(-)(1* 4) (+)(-) (1*1/6)

On using now above stated rules ,

→ -1/3 +4 -1/6

Somewhat rearrange ,

→ 4 -1/3 -1/6

Take (-) as common,

→ 4 - (1/3 +1/6)

<u>Hence Option </u><u>(</u><u>d)</u><u> </u><u>&</u><u> </u><u>(f) </u><u>are</u><u> correct .</u>

I hope this helps.

8 0
2 years ago
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