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3 years ago

What is the probability of choosing a

Mathematics

1 answer:

sladkih [1.3K]3 years ago

3 0

Answer:

Probability of choosing a penny at random is 1/5 or 0.2

Step-by-step explanation:

For many probabilities like this one, you have to add the total of materials or items together, even if they are different colors, shape, coin type, etc.

you have 12 pennies, 19 nickles, 14 dimes, and 15 quarters. This adds up to a total of 60 coins. Twelve of those are pennies.

You have twelve possible pennies to choose at random from a total of 60 in a bag.

So 12/60/ This simplifies to a probability of 1/5. If you divide 1 by 5 than it would be 0.2

The probability of choosing a penny from a bag of coins is 1/5, or 0.2.

Hope this helps

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What are two like fractions with a difference of 1/3

Strike441 [17]

I have no clue sorry

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3 years ago

If n represents a number, then write an expression for a number that is twice n

DaniilM [7]

Answer:

y=n(2)

Step-by-step explanation:

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3 years ago

The equation of the tangent to the curve x^2 = 4y at the point on the curve where x=-2 is?

aleksandrvk [35]

$x^2 = 4y \Rightarrow y= \frac{x^2}{4}$
$\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$

(x = -2) $\frac{dy}{dx} = \frac{x}{2} = \frac{-2}{2} = -1$
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$y-y_1=m(x-x_1)$
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7 0

4 years ago

Read 2 more answers

Find the volume of the sphere.

Ksju [112]

Answer:

$\frac{1}{48} \pi$ units cubed

Step-by-step explanation:

The volume of a sphere is: $\frac{4}{3}\pi r^{3}$

We know the radius r = 1/4, so we just plug this into the equation:

(4/3)* $\pi$ *(1/4)^3 = (4/3)* $\pi$ *(1/64) = $\frac{1}{48} \pi$

So, the answer is $\frac{1}{48} \pi$ units cubed.

4 0

4 years ago

Read 2 more answers

After an extensive advertising campaign, the manager of a company expects the proportion of potential customers that recognize a

lina2011 [118]

Answer:

Step-by-step explanation:

Hello!

The study variable is:

X: number of customers that recognize a new product out of 120.

There are two possible recordable outcomes for this variable, the customer can either "recognize the new product" or " don't recognize the new product". The number of trials is fixed, assuming that each customer is independent of the others and the probability of success is the same for all customers, p= 0.6, then we can say this variable has a binomial distribution.

The sample proportion obtained is:

p'= 54/120= 0.45

Considering that the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample proportion to normal: p' ≈ N(p; $\sqrt{\frac{p(1-p)}{n} }$ )

The other conditions for this approximation are also met: (n*p)≥5 and (n*q)≥5

The probability of getting the calculated sample proportion, or lower is:

P(X≤0.45)= P(Z≤ $\frac{0.45-0.6}{\sqrt{\frac{0.6*0.4}{120} } }$ )= P(Z≤-3.35)= 0.000

This type of problem is for the sample proportion.

I hope this helps!

5 0

3 years ago