Answer:
4.806 g ≤ n ≤ 5.194 where n is the mass of a nickle.
Step-by-step explanation:
The problem tells us the weight can vary by .194 g, this means it can be up to .194 lighter and 194 heavier. Well what value is .194 lighter and .194 heavier? 4.806 and 5.194 respectively. So that means a nickle can be anywhere in that range, which tells us the inequality.
Answer:
x = - 5 , x = 
Step-by-step explanation:
the values of x that make f(x) zero are the zeros
to find the zeros let f(x) = 0 , that is
3x² + 13x - 10 = 0
consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 3 × - 10 = - 30 and sum = + 13
the factors are + 15 and - 2
use these factors to split the x- term
3x² + 15x - 2x - 10 = 0 ( factor the first/second and third/fourth terms )
3x(x + 5) - 2(x + 5) = 0 ← factor out (x + 5) from each term
(x + 5)(3x - 2) = 0
equate each factor to zero and solve for x
x + 5 = 0 ⇒ x = - 5
3x - 2 = 0 ⇒ 3x = 2 ⇒ x = 
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2^n - 3 = 83
<span>2^n = 86 </span>
<span>ln(2^n) = ln(86) </span>
<span>n*ln(2) = ln(86) </span>
<span>n= ln(86)/[ln(2)] (which is the same as "log base 2 of 86") </span>
<span>n= 6.426264755</span>
The probability of it happening once is 1/5, since 10/50 simplifies to 1/5. Because the ticket goes back in the bag, the probability stays constant at 1/5.
For it to happen twice, you multiply the probability of it happening once, twice.
1/5 • 1/5 = 1/25 probability of it being Gary's name twice in a row.
Answer:

Step-by-step explanation:
<u>Given:</u>

<u>Use the Power Rule Law:</u>

<u>Use the Quotient Rule Law:</u>

<u>Use the Product Rule Law:</u>

<u>Simplify:</u>





