I'm completely lost, I need help solving this

bulgar [2K]

Answer:

48oz for $1.39

Step-by-step explanation:

if you divide 48 by 1.39 and you get 34.5323741007

if you divide 32 by .89 you get 35.9550561798

and 34.5323741007 is less than 35.9550561798 and you want the one that costs the least amount of money so you would get the Big Gulp which is 48oz for $1.39

3 0

3 years ago

Find the geometric mean between each pair of numbers. sq root 49 and sq root841

vodomira [7]

Answer:

The geometric mean between each pair of numbers $\bold{\sqrt{49} \text { and } \sqrt{841} \text { is } 14.24}$

Solution:

The Geometric mean between two numbers a and b is given as Geometric mean = $\bold{\sqrt{ab}}$ --- eqn 1

The Geometric Mean is a special type of average where we multiply the numbers together and then take a square root for the numbers.

From question, given that two numbers are $\sqrt{49} and \sqrt{841}$

Hence we can say “a” = $\sqrt{49}$ =7

Similarly “b” = $\sqrt{841}$ = 29

We have to find the geometric mean between “7” and “29”

By using equation 1,

Geometric mean between 7 and 29 = $\sqrt{7 \times 29} = \sqrt{203}$ = 14.24

Hence the geometric mean between $\sqrt{49} \text { and } \sqrt{841} \text { is } 14.24$

8 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

What is the value of tan A?

Fiesta28 [93]

Check the picture below.

4 0

3 years ago

I need help again sorry ;-;

horrorfan [7]

11 units
sin45°=√2 / 2
the length of one leg = √2/2 . 22/√2 = 11

4 0

3 years ago

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