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solniwko [45]
3 years ago
5

Will mark brainliest if correct answer only All,some,or no ​

Mathematics
1 answer:
NeTakaya3 years ago
8 0

Answer:

some

Step-by-step explanation:

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Evaluate 13-0.75w+8x when W=12and x =1/2.
My name is Ann [436]
Hope this will work.

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2 years ago
What are three numbers that have a value between 3/6 and 3/8?​
lisov135 [29]

Answer:

19/50, 9/20, 2/5

Step-by-step explanation:

it's easier to solve if you change 3/6 and 3/8 to decimals which is

0.5 and 0.375 so all you have to do is put in any number between those decimals then change to fractions

6 0
3 years ago
Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.
Tatiana [17]

Answer:

Take the coefficient of ''n'', divide it by 2, and raise it to the second power;

n^2+n+(\frac{1}{2})^2

n^2+n+\frac{1}{4}=0

=(n+\frac{1}{2} )^2

8 0
3 years ago
WILL MARK BRAINLIEST PLEASE HELP I NEED SOME SMART PEOPLE WILL MARK BRAINLIEST!!!
lions [1.4K]

Answer:

The answer is 6\frac{5}{8}

Step-by-step explanation:

-12 pounds + 5\frac{3}{8}

-12+5\frac{3}{8}

=-12+5+\frac{3}{8}

=(-12+5)+\frac{3}{8}

=(-7)+\frac{3}{8\\}

=-6\frac{5}{8}

Mr.Yoo's weight dropped by 6\frac{5}{8}

7 0
2 years ago
Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. If X is the number of defective i
Tcecarenko [31]

Answer:

a) P(X = 0) = 0.5223

b) P(X > 2) = 0.0125

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order that the items are selected is not important, so the combinations formula is used to solve this problem.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this problem, we have that

100 items

6 defective

94 not defective.

a) P{X=0}

None defective.

Desired outcomes

Combinations of 10 from a set of 94. So

D = C_{94,10} = \frac{94!}{10!84!} = 9041256800000

Total outcomes:

Combinations of 10 from a set of 100. So

T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000

P(X = 0)

p = \frac{D}{T} = \frac{9041256800000}{17310309000000} = 0.5223

(b) P(X>2}.

Either two or less are defective, or more than two are defective. The sum of the probabilities of these events is decimal 1. So

P(X \leq 2) + P(X >2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = 0)

P(X = 0) = \frac{D}{T} = \frac{9041256800000}{17310309000000} = 0.5223

P(X = 1)

Desired outcomes

Combinations of 9 from a set of 94(non defecive) and 1 from a set of 6(defective). So

D = C_{84,9}*C_{6,1} = \frac{94!}{9!85!}*\frac{6!}{1!5!} = 6382063700000

Total outcomes:

Combinations of 10 from a set of 100. So

T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000

P(X = 1)

P(X = 1) = \frac{D}{T} = \frac{6382063700000}{17310309000000} = 0.3687

P(X = 2)

Desired outcomes

Combinations of 8 from a set of 94(non defecive) and 2 from a set of 6(defective). So

D = C_{94,8}*C_{6,2} = \frac{94!}{8!86!}*\frac{6!}{2!4!} = 1669726000000

Total outcomes:

Combinations of 10 from a set of 100. So

T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000

P(X = 2)

P(X = 2) = \frac{D}{T} = \frac{1669726000000}{17310309000000} = 0.0965

Finally

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5223 + 0.3687 + 0.0965 = 0.9875

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9875 = 0.0125

4 0
3 years ago
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