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3 years ago

Find two numbers whose difference is 164 and whose product is a minimum. (smaller number) (larger number)

1 answer:

4 0

The lowest possible product would be -6724 given the numbers 82 and -82.

We can find this by setting the first number as x + 164. The other number would have to be simply x since it has to have a 164 difference.

Next we'll multiply the numbers together.
x(x+164)
x^2 + 164x

Now we want to minimize this as much as possible, so we'll find the vertex of this quadratic graph. You can do this by finding the x value as -b/2a, where b is the number attached to x and a is the number attached to x^2

-b/2a = -164/2(1) = -164/2 = -82

So we know one of the values is -82. We can plug that into the equation to find the second.

x + 164
-82 + 164
82

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3 years ago

The number of tourists at the beach per weekend in July was 55,000, In November, the number of tourists per weekend was 18,000.

Ganezh [65]

Answer:

The percentage change from July to November is 67.27 %

Step-by-step explanation:

Given as :

The number of tourists at the beach per weekend in the month of July = $x_1$ = 55,000

The number of tourists at the beach per weekend in the month of November = $x_2$ = 18,000

Let the percentage change from July to November = A %

Or, % decrease change = $\dfrac{\tetxtrm new value - \textrm old value}{\textrm old value}$ × 100

So , A % = $\dfrac{\tetxtrm [tex]x_2$ - \textrm $x_1$ }{\textrm $x_1$ }[/tex] × 100

or, A % = $\frac{55000 - 18000}{55000}$ × 100

Or, A % = $\frac{37000}{55000}$ × 100

Or, A = 67.27 %

So percentage change between two months = 67.27 %

Hence The percentage change from July to November is 67.27 % Answer

6 0

3 years 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