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3 years ago

14

I need help solving them and quick cause I don’t have a lot of time

1 answer:

riadik2000 [5.3K]3 years ago

7 0

A: x = 2.4
b: m = 9.6
c: c = 2 5/12
d: m = 3.6
e: c = 6.6

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Kelly went to a hockey game with her family. They bought 4 bags of cotton candy and 2 souvenir cups and spent $36. Marissa's fam

erik [133]

4b+2c=36, solve for b, subtract 2c from both sides

4b=36-2c divide both sides by 4

b=9-0.5c

Then we are told that:

7b+3c=59, using b=9-0.5c we get:

7(9-0.5c)+3c=59 perform indicated multiplication on left side

63-3.5c+3c=59 combine like terms on left side

63-0.5c=59 subtract 63 from both sides

-0.5c=-4 divide both sides by 0.5

c=8

So each souvenir cup costs $8.

(and each bag of cotton candy is $5)

3 0

3 years ago

How many times can 8 go into 329

tankabanditka [31]

8 can go into 329 wholly 41 times. However, there is a remainder of 0.125.

5 0

3 years ago

Please Help!! what is 4+4?

Artist 52 [7]

Answer:

8

Step-by-step explanation:

4 0

3 years ago

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A man has n keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he r

schepotkina [342]

Answer:

The value is $P(3) = \frac{1}{n}$

Step-by-step explanation:

From the question we are told that

The number of keys is n

The number of keys remaining after the first key is chosen is n-1

The probability that he gains entrance with the first key he selects is $\frac{1}{n}$

Generally the probability that the first key does not open the door is

$p(F_1) = 1 - \frac{1}{n} = \frac{n-1}{n}$

Generally the number of keys remaining after the second key is chosen is

n-2

Generally the probability that he gains entrance with the second key he selects is

$\frac{1}{n-1}$

Generally the probability that the second key does not opens the door is

$P(F_2) = 1- \frac{1}{n-1} = \frac{n-2}{n-1}$

Generally the probability that he gains entrance with the third key he selects is

$\frac{1}{n-2}$

Generally the probability that the door opens with the third key

$P(3) = p(F_1) * P(F_2) * \frac{1}{n-2}$

=> $P(3) = \frac{n-1}{n} * \frac{n-2}{n-1} * \frac{n-2}{n-1}$

=> $P(3) = \frac{1}{n}$

Note :

All the outcome of the events are independent

7 0

4 years ago

What is x in this 5x-6 = -6

enot [183]

Answer:

x=0

Step-by-step explanation:

The x has to be by themselves so get rid of the 5 and -6 - -6 equals 0

7 0

3 years ago

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