Please help. 50 points on the line

X +5+11x=12x+Y It says find the value of y that makes each equation true for all values of

sladkih [1.3K]

X + 5 + 11x = 12x + y

Simplify: 12x + 5 = 12x + y (We are adding x and 11x on the left side)

Subtract 12x from each side makes each zero.

5 = y

So we can plug in and test. I'm picking two random numbers to plug in for x. 10, and 82

x = 10, y = 5
10 + 5 + 11(10) = 12(10) + 5
125 = 125

x = 82, y = 5
82 + 5 + 11(82) = 12(82) + 5
989 = 989

So we verified y = 5

3 0

3 years ago

The original price of a bicycle is $130. The sale price is 60% of the original price. What is the sale price of the bicycle?

djverab [1.8K]

Answer:

$78

Step-by-step explanation:

To find a portion of anything, you just multiply your first item by the equivalent decimal of your percent. Let's put it into action in this problem:

If your original price of your bike is $130, and you are trying to find the sale price, which is 60% (aka 0.6) of your original price. Now, like I previously mentioned, to find the portion of your price, you just want to multiply your OG price (130) by the decimal equivalent of your sale, (0.6). So 130*0.6=78

3 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