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2 years ago

A collection of 77 quarters and dimes is worth $12.50. How many quarters and dimes are there? Set it up plz

1 answer:

mihalych1998 [28]2 years ago

4 0

So let's say quarters = x and dimes = y. You'd then put x + y = 77 (for a total of 77 coins). And your second equation would be .25x + .10y = 12.50 (.25 for the total value a quarter holds and .10 for a total value a dime holds). Then you'd do matrices on a calculator.

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A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

1 year ago

Jamie took 20 pieces of same-sized colored paper and put them in a hat. Four pieces were red, three pieces were blue, and the re

shtirl [24]

The chance she will pull out a red paper is 20%.

3 0

1 year ago

How do you simplify (25/a -a l) / (5+a)

andrey2020 [161]

Simplify the following:

(25/a - a l)/(a + 5)

Put each term in 25/a - a l over the common denominator a: 25/a - a l = 25/a - (a^2 l)/a:

(25/a - (a^2 l)/a)/(a + 5)

25/a - (a^2 l)/a = (25 - a^2 l)/a:

Answer: ((25 - a^2 l)/a)/(a + 5)

7 0

2 years ago

In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t

victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that he or she would choose the pair of socks in the center position is $p =\frac{1}{5}$

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5

The mean is $\mu = 20$

The standard deviation is $\sigma=4$

The probability, $P =0.0002$

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability $Pe = 0.0003$

The probabilities $P \approx Pe$

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

$p =\frac{1}{5}$

$=0.2$

The mean of this distribution is mathematical represented as

$\mu = np$

substituting the value

$\mu = 100 * 0.2$

$\mu = 20$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np (1-p)}$

substituting the value

$= \sqrt{100 * 0,2 (1-0.2)}$

$\sigma=4$

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

$P=P(X \ge 34 )$

By standardizing the normal approximation we have that

$P(X \ge 34) \approx P(Z \ge z)$

Now z is mathematically evaluated as

$z = \frac{x-\mu}{\sigma }$

Substituting values

$z = \frac{34-20}{4}$

$=3.5$

So using the z table the $P(Z \ge 3.5)$ is 0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

6 0

2 years ago

The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a

maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

$\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}$

$\text{P(green swordtail)}=\dfrac{80}{140}$

$\text{P(green swordtail)}=\dfrac{4}{7}$

Therefore, the probability that the selected fish is a green swordtail is $\dfrac{4}{7}.$

B. We have to find the probability that the selected fish is male.

$\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}$

$\text{P(Male fish)}=\dfrac{36+24}{140}$

$\text{P(Male fish)}=\dfrac{60}{140}$

$\text{P(Male fish)}=\dfrac{3}{7}$

Therefore, the probability that the selected fish is a male, is $\dfrac{3}{7}.$

C. We have to find the probability that the selected fish is a male green swordtail.

$\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}$

$\text{P(Male green swordtail)}=\dfrac{36}{140}$

$\text{P(Male green swordtail)}=\dfrac{9}{35}$

Therefore, probability that the selected fish is a male green swordtail is $\dfrac{9}{35}.$

We have to find the probability that the selected fish is either a male or a green swordtail.

$\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}$

$\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}$

$\text{P(Male or green swordtail)}=\dfrac{96}{140}$

$\text{P(Male or green swordtail)}=\dfrac{24}{35}$

Therefore, the probability the selected fish is either a male or a green swordtail is $\dfrac{24}{35}.$

4 0

2 years ago