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nordsb [41]
3 years ago
9

Help me please sjsjdjs it’s due in a few minutes

Mathematics
2 answers:
Orlov [11]3 years ago
8 0

Answer:

The answer is the 3rd dot.

Step-by-step explanation:

hope this helps

Citrus2011 [14]3 years ago
6 0

Answer:

The one right below 8

Step-by-step explanation:

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What is the solution to the equation? 13 3/4+x=7 1/4
oksano4ka [1.4K]

Answer:

x=-13/2

Step-by-step explanation:

13 3/4=55/4

7 1/4=29/4

---------------------

55/4+x=29/4

x=29/4-55/4

x=-26/4

simplify

x=-13/2

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3 years ago
Find the total surface area of this triangular prism 8cm 30cm 24cm 7cm
Alenkinab [10]

Answer:

61

Step-by-step explanation:

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3 years ago
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
Solve for x<br> -6x+14&lt; -28 or 9x+15≤−12
Free_Kalibri [48]

- 3 \geqslant x > 6

Step-by-step explanation:

....

8 0
3 years ago
Help me solve for k please anyone
kobusy [5.1K]

Answer:

Rounding the the nearest tenth then it is 3.5

Step-by-step explanation:

7 0
3 years ago
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