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3 years ago

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A coon bank contains only quarters and dimes.There are 25 coins in all.Seven of the coins are dimes.What fraction of the couns a

re quarters?

2 answers:

iris [78.8K]3 years ago

4 0

Answer:18/25

Step-by-step explanation:

horrorfan [7]3 years ago

3 0

I agree 18/25, which will be 0.72. Hope it helped

You might be interested in

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

3 years ago

Use technology or a z-score table to answer the question.

Alik [6]

Answer:

The second choice: Approximately $65.2\%$ of the pretzel bags here will contain between 225 and 245 pretzels.

Step-by-step explanation:

This explanation uses a z-score table where each $z$ entry has two decimal places.

Let $\mu$ represent the mean of a normal distribution of variable $X$ . Let $\sigma$ be the standard deviation of the distribution. The z-score for the observation $x$ would be:

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

In this question,

$\mu = 240$ .
$\sigma = 9.3$ .

Calculate the z-score for $x_1 = 225$ and $x_2 = 245$ . Keep in mind that each entry in the z-score table here has two decimal places. Hence, round the results below so that each contains at least two decimal places.

$\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}$ .

$\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}$ .

The question is asking for the probability $P(225 \le X \le 245)$ (where $X$ is between two values.) In this case, that's the same as $P(-1.61 \le Z \le 0.54)$ .

Keep in mind that the probabilities on many z-table correspond to probability of $P(Z \le z)$ (where $Z$ is no greater than one value.) Therefore, apply the identity $P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)$ to rewrite $P(-1.61 \le Z \le 0.54)$ as the difference between two probabilities:

$P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61)$ .

Look up the z-table for $P(Z \le 0.54)$ and $P(Z \le -1.61)$ :

$P(Z \le 0.54)\approx 0.70540$ .
$P(Z \le -1.61) \approx 0.05370$ .

$\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}$ .

3 0

3 years ago

What is 3 and 5? And show how to solve plz

nataly862011 [7]

Answer:

3. 9n + 6 = 78

5. see below

Step-by-step explanation:

3.

Let the number of adults be n. Each adult ticket costs $9, so the cost of n adult tickets is 9n. There is only one child, and a child's ticket is $6, so the cost of the child's ticket is 6. Adding the costs, you get 9n + 6. The total cost is $78, but the expression representing the total cost is 9n + 6, so 9n + 6 must equal 78. That is the equation.

9n + 6 = 78

5.

Sam bought equal gifts for his two children. The two gifts had the same price, x. He received a discount of $125. How much did each of the two gifts cost before the discount?

5 0

3 years ago

Please can someone do this for me? thanks alot x

Nimfa-mama [501]

Answer:

<em>Learning Objective</em>: Use the mean and range to <u>compare</u> the spread and averages.

a) Sarah Mean = 70.9, Katie Mean = 78.9

b) Sarah Range = 15, Katie Range = 25

c) Sentences

- The <u>mean</u> shows, on average, Sarah was 8 seconds quicker than Katie.

- The <u>range</u> shows that Sarah was more consistent with her times. Katie had a larger spread of times which shows she was less consistent.

Step-by-step explanation:

Given:

Sarah's Data Set:

79, 70, 68, 75, 69, 64, 69, 75, 64, 76

Katie's Data Set:

79, 79, 76, 81, 89, 76, 64, 85, 82, 78

To Find:

The mean and range of both data sets separately.

Work:

The mean of a data set is commonly known as the average. You find the mean by taking the sum of all the data values and dividing that sum by the total number of data values. The formula for the mean of a population is $\mu = \frac{{\sum}x}{N}$ .

The formula for the mean of a sample is $\bar{x} = \frac{{\sum}x}{n}$ .

Both of these formulas use the same mathematical process: find the sum of the data values and divide by the total. For the data values given above, the solution is: $\frac{709}{10} = 70.9$

Therefore, the mean of Sarah's data set is 70.9.

For Katie's Data Set, $\frac{789}{10} = 78.9$ .

Therefore, the mean of Katie's data set is 78.9.

To fill in the summary, we have to use the data we just calculated.

That summary can be found in the answer section.

4 0

1 year ago

Read 2 more answers

400,000,000,000 90,000,000,000 6,000,000,000 60,000,000 300,000 40,000 2,000 800 10 1 in standard form

Greeley [361]

Perhaps you want the sum: 496,060,342,811.

7 0

3 years ago