All categories

3 years ago

GIVING BRAINLIEST ONCE AGAIN

Mathematics

2 answers:

german3 years ago

4 0

Answer:

5:9 = 25:45

Step-by-step explanation:

Andru [333]3 years ago

3 0

Answer:

Step-by-step explanation:

divide 25 by 5 to see how they got that, which is 5. then multiply 9 by 5 bc you have to do the same thing, which is 45.

You might be interested in

ElenaW [278]

Hold on let me go do my research I’ll brb .

4 0

3 years ago

2.6.5 A plant physiologist grew birch seedlings in the green-house and measured the ATP content of their roots. (See Example 1.1

Ilya [14]

Answer:

$(a)\ \bar x = 1.19$

$(b)\ \sigma_x = 0.18$

Step-by-step explanation:

Given

$n = 4$

$x: 1.45\ 1.19\ 1.05\ 1.07$

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{1.45 + 1.19 + 1.05 + 1.07}{4}$

$\bar x = \frac{4.76}{4}$

$\bar x = 1.19$

Solving (b): The standard deviation

This is calculated as:

$\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n - 1}}$

So, we have:

$\sigma_x = \sqrt{\frac{(1.45 -1.19)^2 + (1.19 -1.19)^2 + (1.05 -1.19)^2 + (1.07 -1.19)^2}{4 - 1}}$

$\sigma_x = \sqrt{\frac{(0.1016}{3}}$

$\sigma_x = \sqrt{0.033867}$

$\sigma_x = 0.18$

4 0

3 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

3 years ago

PLEASE HELP WITH THIS EQUATION!!! WILL GIVE BRAINLIEST AND RATING!!!!!!!

GuDViN [60]

Answer:

$-8x+6$