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3 years ago

Order these fractions from least to greatest 7/8 4/6 3/4

Mathematics

1 answer:

barxatty [35]3 years ago

8 0

Answer:

4/6 , 3/4, 7/8

Step-by-step explanation:

7/8 4/6 3/4

The common denominator is 24

7/8*3/3 = 21/24

4/6* 4/4 = 16/24

3/4 * 6/6 = 18/24

From smallest to largest

16/24, 18/24, 21/24

4/6 , 3/4, 7/8

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A home gardener estimates that 18 apple trees will produce an average yield of 80 apples per tree. But because of the size of th

aliina [53]

19 trees should be planted to maximize the total

<h3>How many trees should be planted to maximize the total</h3>

From the question, we have the following parameters:

Number of apples, x = 18

Yield, f(x) = 80 per tree

When the number of apple trees is increased (say by x).

We have:

Trees = 18 + x

The yield decreases by four apples per tree.

So, we have

Yield = 80 - 4x

So, the profit function is

P(x) = Apples * Yield

This gives

P(x) = (18 + x) *(80 - 4x)

Expand the bracket

P(x) = 1440 - 72x + 80x - 4x^2

Differentiate the function

P'(x) = 0 - 72 + 80 - 8x

Evaluate the like terms

P'(x) = 8 - 8x

Set P'(x) to 0

8 - 8x = 0

Divide through by 8

1 - x = 0

Solve for x

x = 1

Recall that:

Trees = 18 + x

So, we have

Trees = 18 + 1

Evaluate

Trees = 19

Hence, 19 trees should be planted to maximize the total

Read more about quadratic functions at:

brainly.com/question/12120831

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4 0

1 year ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago

A school debate team has 4 girls and 6 boys. A total of 3 of the team members will be chosen to participate in the district deba

lys-0071 [83]

Answer:

I’m pretty sure the answer is A.

Hope this helps!

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3 years ago

What is the discriminant of 3x^2 - 10 = -2?

Tatiana [17]

The formula for the discriminant is b^2-4ac. The formula you wote is in the form of
ax^2+c=0, so first, you need to bring that -2 to the left. When you do that, you get (the original equation would be ax^2+bx+c but you have no bx.)

3x^2-8

So, since there's no b, it would be
b^2-4ac
0^2-4(3)(-8
-12 x -8
96

7 0

3 years ago

Write two expressions that show that decreasing a quantity by 10% is the same as multiplying that quantity by 0.9

Contact [7]

Step-by-step explanation:

x -0.1x = x(1-0.1) = 0.9x

3 0

3 years ago