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Svet_ta [14]
3 years ago
8

How many times can 14 go into 8574?

Mathematics
1 answer:
bearhunter [10]3 years ago
8 0
612.4285714285714 is the answer
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1 + 3 - (8 * 334 * 123) = x<br><br> x = ?
il63 [147K]

Answer:

if, 1 + 3 - (8 * 334 * 123) = x

then, x = -328652

8 0
3 years ago
Need your guyzes halp, no false answers and I will give you most of my points
Alex17521 [72]

Answer$32

Step-by-step explanation:

125 divided by 2 = 31.25 +00.05+32

8 0
2 years ago
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Add 3/14+4/14 simply your answer
Mila [183]

Answer:

\Large \boxed{\frac{1}{2}}

Step-by-step explanation:

\displaystyle \frac{3}{14} +\frac{4}{14}

Add the fractions since the denominators are the same.

\displaystyle \frac{3+4}{14}

\displaystyle \frac{7}{14}

Simplifying the fraction.

\displaystyle \frac{7(1)}{7(2)}

\displaystyle \frac{1}{2}

5 0
3 years ago
The function f(t)= 5 tan 2 t, does not have an amplitude and has a period of π.
trasher [3.6K]

ANSWER

False

EXPLANATION

The tangent function has no amplitude because it is not bounded.

The given tangent function is

f(t) = 5 \tan(2t)

This is of the form

f(t)=a tan(bt)

The period is given by

T =  \frac{\pi}{ |b| }

T =  \frac{\pi}{ |2| }  =  \frac{\pi}{2}

The first statement is true but the second is false.

Hence the whole statement is false.

8 0
3 years ago
Read 2 more answers
There are 5 pens in a container on your desk. Among them, 3 will write well but 2 have defective ink cartridges. You will select
Norma-Jean [14]

Answer:

Kindly check explanation

Step-by-step explanation:

Total Number of pens = 5

Number of defective pens = 2

Number of non-defective pens = 3

A.) number of defective pens selected :

X : {0, 1, 2}

It is possible that no defective pen will be selected ; 1 defective will be chosen or both pens are defective.

2.)

Defective as Success (since selecting a defective pen is the point of interest.

3.)

Since selection is done without replacement

Probability of success per selection is different for each selection ;

Number of defective = 2

Number of observations = 5

P(success on first selection) = 2/5

P(success on second selection) = 1/4

Hence, X is not well approximated by a binomial random variable.

4.) if selection is done with replacement ; then then the probability of success per selection will be the Same for each selection made. Hence, X will be well approximated by a binomial Random variable.

5.) If sampling is done without replacement, then the the hypergeometric function will be a more effective approximation.

7 0
3 years ago
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