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alexgriva [62]
3 years ago
12

A box filled with brass connectors weighs 5 pounds and 4 ounces. The box weighs 8 ounces when empty and each connector weighs 0.

25 ounces. How many connectors should be in the box?
A. 294
B. 304
C. 336
D. 368
E. 397
Mathematics
2 answers:
anyanavicka [17]3 years ago
8 0
A ........hvxxjizjzf if if
mario62 [17]3 years ago
3 0
B 304. Total weight without box is 4lbs 12 Oz. Or 76 Oz, 76x4 connectors per Oz = 304
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$80 times 0.25 equals 20
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For 1 bath its 2 liter so how many liers is it for 2 baths ​
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4

Step-by-step explanation:

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At a certain bakery, the price of each doughnut is $1.50. Let the random variable D represent the number of doughnuts a typical
Solnce55 [7]

Using <em>statistical concepts</em>, it is found that the standard deviation of the probability distribution is of

s = 1.5(1.9) = 2.85

When a constant is multiplied to each data in a variable, both the <u>mean and the standard deviation</u> are multiplied by the constant.

In this problem, the standard deviation, which is the square root of the variance, is of:

\sigma = \sqrt{\sigma^2} = \sqrt{3.6} = 1.9

The price of each is of $1.5, hence:

s = 1.5(1.9) = 2.85

A similar problem is given at brainly.com/question/25718546

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B5x%20%2B%2050%7D%7Bx%20%2B%205%7D%20.%20%5Cfrac%7B1%7D%7Bx%20%2B%2010%7D%20" id="
bezimeni [28]
Hi, I'd be glad to help you with this equation. 

Your answer is (5/ x + 3) 

Hope this helps. 
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8 0
4 years ago
Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p
scZoUnD [109]

Answer:

The probability that they will teach different courses is \frac{2}{3}.

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here <em>x</em> represents the course taught by the first part-time instructor and <em>y</em> represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

Compute the probability that they will teach different courses as follows:

P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}

Thus, the probability that they will teach different courses is \frac{2}{3}.

3 0
3 years ago
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