Describe two different strategies you could use to solve 16+35+24+14

15. The graph of a linear function, f(x), is shown below. If the line is translated 2 units

Leto [7]

Answer: fx-2 i believe since translating down would subtract two

Step-by-step explanation:

4 0

2 years ago

Plz some one help me

sveta [45]

9 games

You can get this by multiplying each percentage by the total number of games to see what they have to expect.

.55*180 = 99 games (red team)

.60*180 = 108 games (blue team)

Now subtract the blue team expectation from the red team.

108 - 99 = 9 games.

6 0

3 years ago

Is a letter that is used to represent a number.

Novay_Z [31]

Answer:

variable

Step-by-step explanation:

A variable is a American, Latin, Greek etc letter used to represent an unknown.

7 0

2 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

Does it pay to ask for a raise? A national survey of heads of households showed the percentage of those who asked for a raise an

zvonat [6]

Answer:

Cross probability

Step-by-step explanation:

The type of probabilities remains to be determined. That is the question itself. The same problem is found in the book "Understanding basic statistics 7th edition of Brase".

I will assume that the three probabilities to find are:

a) p (woman asked for a raise)

b) p (received raise | asked for one)

c) p (received raise and asked for one)

So,

a) p(woman asked for a raise)

It is given in the statement, so it is simple statistics:

$P_{WomA} =25%$

b) p(received raise | asked for one)

It is also found in the statement, however, now the formula would be given by:

p(received raise | asked for one) = p(received raise and asked for one) / p(asked for one)

$P_{WomB}=46%$

c) p(received raise and asked for one)

For this case the formula is given differently and it is necessary to re-adjust formula B, as follows:

p(received raise and asked for one) = p(received raise | asked for one) * p(asked for one)

So,

$P_{WomC} = 0.46*0.25 = 11.5%$

p(received raise and asked for one) = 11.5%

7 0

3 years ago