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Oksi-84 [34.3K]

3 years ago

5

Esteban has a big jar of change in his room. He has 600 coins total, and 240 of them are pennies. What percentage of the coins a

re pennies?

1 answer:

mina [271]3 years ago

3 0

Answer:

40%

Step-by-step explanation:

The fraction of this problem would be 240/600 and because a percent is over 100 you would have to solve for that. To make 600 to be 100 you would divide it by 6, and do it for both sides.

$\frac{240}{600}$ ÷ $6$ = 40/100

So the percent would be 40%

Hope that helps and have a great day!

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Rice weighing 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?

QveST [7]

Answer:

The answer is 15 ounces

Step-by-step explanation:

33/4 ÷ 4 pounds.

(4 × 3 + 3)/4 ÷ 4 pounds.

15/4 ÷ 4 pounds.

15/4 × 1/4 pounds.

15/16 pounds.

Now we know that, 1 pound = 16 ounces.

So, 15/16 pounds = 15/16 × 16 ounces = 15 ounces.

Thus, The answer is 15 ounces

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3 years ago

Can you help please?

lina2011 [118]

-5/2

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2 years ago

Read 2 more answers

What ratios are equivalent to 12:3

nlexa [21]

Answer:

4:1

Step-by-step explanation:

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3 years ago

Read 2 more answers

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

erma4kov [3.2K]

Answer:

a) $n = 9604$

b) n = 381

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?

When we do not know the proportion, we use $\pi = 0.5$ , which is when we are going to need the largest sample size.

The sample size is n when M = 0.01.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.01\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.01})^{2}$

$n = 9604$

b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Now we have that $\pi = 0.99$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.99*0.01}{n}}$

$0.01\sqrt{n} = 1.96*\sqrt{0.99*0.01}$

$\sqrt{n} = \frac{1.96*\sqrt{0.99*0.01}}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.99*0.01}}{0.01})^{2}$

$n = 380.3$

Rouding up

n = 381

7 0

3 years ago

6. (4.2.12) Of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired. a. F

nata0808 [166]

Answer:

(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

Step-by-step explanation:

We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.

Let Probability that item are defective = P(D) = 0.20

Also, R = event of item being repaired

Probability of items being repaired from the given defective items = P(R/D) = 0.60

<em>So, Probability of items not being repaired from the given defective items = P(R'/D) = 1 - P(R/D) = 1 - 0.60 = 0.40 </em>

(a) Probability that a randomly chosen item is defective and cannot be repaired = Probability of items being defective $\times$ Probability of items not being repaired from the given defective items

= 0.20 $\times$ 0.40 = 0.08 or 8%

So, probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 20 items

r = number of success = exactly 2

p = probability of success which in our question is % of randomly

chosen item to be defective and cannot be repaired, i.e; 8%

<em>LET X = Number of items that are defective and cannot be repaired</em>

So, it means X ~ $Binom(n=20, p=0.08)$

Now, Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is given by = P(X = 2)

P(X = 2) = $\binom{20}{2} \times 0.08^{2} \times (1-0.08)^{20-2}$

= $190 \times 0.08^{2} \times 0.92^{18}$

= 0.2711

<em>Therefore, probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is </em><em>0.2711.</em>

6 0

3 years ago