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Sergio039 [100]
3 years ago
13

a gym charges a $50 activation fee and $17 per month for a membership.if you spend $356, for how many months do you have a gym m

embership
Mathematics
2 answers:
nydimaria [60]3 years ago
8 0
356 - 50 = 306

306 / 17 = 18

18 months of the gym membership
Maslowich3 years ago
8 0
356 = 50 + 17m
306 = 17m
18 = m
18 months
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The value 191.5638 rounded to the nearest hundredth is<br> 191.564<br> 191.5600<br> 191.56<br> 200
Vesna [10]

Answer:

191.56

Step-by-step explanation:

Given

191.5638

Required

Approximate

Approximating to the nearest hundredth implies that, only to digits after the decimal point is needed

This is:

191._ _

The third digit after the decimal point will approximated using the approximation rule;

Since, the third digit after the decimal point (3) is not up to 5, it'll be approximated to 0 and added to the second digit (6)

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191.5 _

The next digit is calculated as 6 + 0 = 6

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6 0
3 years ago
Can anyone help??? I’m stuck
sukhopar [10]

Answer:

C)

Step-by-step explanation:

If you take a look at each of the divided lines, 4 out of the 5 segments are filled, representing 4/5. Since there are 6 of lines, the answer is 6\times \dfrac{4}{5}, or choice C. Hope this helps!

4 0
3 years ago
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You are to play the following game of cards. Cards are worth their face value, Jacks, Queens and Kings are also worth 10 and Ace
ryzh [129]

Answer:

$82.875

Step-by-step explanation:

From the given information:

Assume F is used to denote the two cards;

If there are four aces among 52 playing cards, the chance of selecting the first ace is =  \dfrac{4}{52}

After selecting the first ace, we have only 3 aces remaining and a total of 51 playing cards. Thus, the chance of selecting the second ace will be \dfrac{3}{51}

By applying product rule, we can determine the chance of selecting two aces without replacement as follows:

i.e.

P(F=22)=\dfrac{4}{52}\times \dfrac{3}{51}

= \dfrac{1}{221}

The probability of getting one ace, one face is:

P(F=21) =( \dfrac{4}{52}\times \dfrac{12}{51})+( \dfrac{12}{52}\times \dfrac{4}{51})

P(F=21) = \dfrac{4}{221}\times\dfrac{4}{221}

P(F=21) = \dfrac{8}{221}

Since there are 4 aces, 4 nine, and 12 faces in a card deck

The probability of getting one ace, one nine, or two faces now will be:

P(F=20) = (\dfrac{4}{52} \times \dfrac{4}{51})+ (\dfrac{4}{52} \times \dfrac{4}{51}) + (\dfrac{12}{52} \times \dfrac{11}{51})

P(F=20) = (\dfrac{53}{663}  )

Now, the probability  of at least 20 now is:

\text{P(F at least 20)} = \dfrac{1}{221}+\dfrac{8}{221}+\dfrac{53}{663}

\text{P(F at least 20)} = \dfrac{80}{663}

If H represents the amount of prize of the expected winnings:

Then;

(H - 10) (\dfrac{80}{663}) + (-10)(\dfrac{663-80}{663}) = 0

\dfrac{80(H-10)}{663}-\dfrac{5830}{663}=0

\dfrac{80(H-10)}{663}=\dfrac{5830}{663}

80H - 800 = 5830

80H = 5830 +800

80H = 6630

H = 6630/80

H = $82.875

The prize should be $82.875 to make a winning positive.

4 0
3 years ago
(PLEASE SOLVE ASAP EXPLAIN ALSO)
adelina 88 [10]

Answer:

3.81

Step-by-step explanation:

-      +     -

-      +     -

-      +      -

If you were to go all ways on this chart↑ the rules would always be the same.

A negative plus a negative is a negative.

A negative plus a positive is a negative.

A positive plus a postitive is a positive.

5 0
3 years ago
Add the following integers {3 +(-7)}+(-6)​
Travka [436]

Answer:

-10

Step-by-step explanation:

3 0
3 years ago
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