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3 years ago

Using the least number of coins how many quarters, dimes, nickels, and pennies to make $1.74

Mathematics

1 answer:

SpyIntel [72]3 years ago

4 0

Let's use our largest amount (quarter) to see how many can fit evenly into 1.74.

1.74/0.25 = 6.96
The most quarters that can fit into $1.74 is 6.

---Multiply $0.25 by 6 = $1.50

---Subtract $1.50 from $1.74 = $0.24

---We need to use the rest of the coins to fill up $0.24.

Now let's use our second largest amount (dime) to see how many can fit evenly into 0.24.

0.24/0.10 = 2.4
The most dimes that can fit into $0.24 is 2.

---Multiply $0.10 by 2 = $0.20

---Subtract $0.20 from $0.24 = $0.04

No nickels can fit into $0.04, because they are worth $0.05.

We can use our least amount (pennies) to fill in the $0.04 remaining.

Your answer is 6 quarters, 2 dimes, and 4 pennies.

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Simplify the expression.<br> (x 3/2)6

Sliva [168]

For this case we must simplify the following expression:

$(x ^ {\frac {3} {2}}) ^ 6$

We have that by definition of properties of powers that is fulfilled:

$(a ^ n) ^ m = a ^ {n * m}$

Then, rewriting the expression:

$x ^ {\frac {3 * 6} {2}} =\\x ^ {\frac {18} {2}} =\\x ^ 9$

Answer:

$(x^{\frac{3}{2}})^6=x^9$

6 0

3 years ago

A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a

netineya [11]

Answer:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=19, p=0.3)$