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alisha [4.7K]
3 years ago
6

Can you walk me thru this 2 •(3(5+2)-1)​

Mathematics
1 answer:
kobusy [5.1K]3 years ago
5 0

Answer:

40

Step-by-step explanation:

2 •(3(5+2)-1)​

2 ·(3(7)-1)

2·(21-1)

2·20

=40

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I need help pls:) and if u need help just tell me
Mariulka [41]
The answer is 28

84/3 =28
8 0
2 years ago
Mark sold x pens for $2 each. He received less than $80 from the sale. Find the greatest possible number of pens he sold
Crank
$80-$2= $78 is the number less than $80, since the pens cost $2

78/2= 39 pens is the greatest number of pens sold

hope this helps
6 0
3 years ago
Read 2 more answers
Which is the correct simplification:<br> giving brainliest
gladu [14]

Answer:

The answer is letter E.........

5 0
2 years ago
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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What
Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

a) What is the probability that none of the questions are essay?

P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

8 0
3 years ago
Seven less than the product of a number and three equal to eleven
Anon25 [30]

Answer:

7-(3x)=11

Step-by-step explanation:

4 0
3 years ago
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