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Anettt [7]
3 years ago
5

David has nickels and y dimes. He has at least 24 coins worth at most $1.80

Mathematics
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

Step-by-step explanation:

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Pls help me with this multiple choice !
lana66690 [7]

Answer:

D

Step-by-step explanation:

\sin(60)  =  \frac{6 \sqrt{3} }{x}  \\ x =  \frac{6 \sqrt{3} }{ \sin(60) }  = 12 \\  \tan(60)  =  \frac{6 \sqrt{3} }{y}   \\ y =  \frac{6 \sqrt{3} }{ \tan(60) }  =   6

8 0
3 years ago
Molly has 190 dollars in her bank account and spend 10 dollars every trip write an equation for molly
leonid [27]
10xt=190
T represents the amount of trips and 10 represents the amount of money she spends each trip. Since she has $190 already it will be on the other side of the equation after the equal sign
3 0
3 years ago
Divide -1.53 ÷ 0.17 enter your answer in the box PLEASE HELLPPP
marshall27 [118]

Answer:

-9

Step-by-step explanation:

-1.53  ÷  0.17 =‬ -9

check work:

-9 x 0.17 = -1.53

6 0
3 years ago
Read 2 more answers
Slope=1÷5; y intercep = -5
sertanlavr [38]
The equation is
y =  \frac{1}{5}  x  - 5
3 0
4 years ago
A certain restaurant always overbooks. If possible. At 7:00 pm the restaurant can seat 50 parties, but takes reservations for 53
oee [108]

Answer:

The probability that the restaurant can accommodate all the customers who do show up is 0.3564.

Step-by-step explanation:

The information provided are:

  • At 7:00 pm the restaurant can seat 50 parties, but takes reservations for 53.
  • If the probability of a party not showing up is 0.04.
  • Assuming independence.

Let <em>X</em> denote the number of parties that showed up.

The random variable X follows a Binomial distribution with parameters <em>n</em> = 53 and <em>p</em> = 0.96.

As there are only 50 sets available, the restaurant can accommodate all the customers who do show up if and only if 50 or less customers showed up.

Compute the probability that the restaurant can accommodate all the customers who do show up as follows:

P(X\leq 50)=1-P(X>50)\\=1-P(X=51)-P(X=52)-P(X=53)\\=1-[{53\choose 51}(0.96)^{51}(0.04)^{53-51}]-[{53\choose 52}(0.96)^{52}(0.04)^{53-52}]\\-[{53\choose 53}(0.96)^{53}(0.04)^{53-53}]\\=1-0.27492-0.25377-0.11491\\=0.3564

Thus, the probability that the restaurant can accommodate all the customers who do show up is 0.3564.

7 0
3 years ago
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