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2 years ago

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Winnie wrote the following riddle.I am a number between 60 and 100.My ones digit is two less than my tens digit.I am a prime num

ber.

1 answer:

Kazeer [188]2 years ago

5 0

Answer:

79

Step-by-step explanation:

60<xy<100

y -2 =x

if it is prime

y cannot be even

so y could be 1,3,5,7,9

x would then be 9,1,3,5,7

but to be between 60 and 100

x must be 7 or 9

so the number must be 79 or 57

57 is not prime so

the number is 79

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2 years ago

There are 15 students in the 8th grade. The students are randomly placed into three different algebra classes of 5 students each

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Answer:

The probability is $\frac{6}{91}$ ≅ $0.0659$

Step-by-step explanation:

Let's analyze the question.

There are 15 students in the 8th grade.

The students are randomly placed into three different algebra classes of 5 students each.

We are looking for the probability that Trevor, Terry and Evan will be in the same algebra class.

One possible way to solve this question is to think about the product probability rule.

We can use it because we are in an equiprobable space. (And also the events are independent).

Let's set for example a class for Evan.

The probability that Evan will be in a class is $1$

Then for Terry there are $4$ places out of $14$ that puts Terry in the Evan's class.

We write $1.\frac{4}{14}$

Finally for Trevor there are $3$ places out of the remaining $13$ that puts Trevor in the same class with Evan and Terry.

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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 <em>sample mean</em>. The distribution for <em>sample means</em> follows a <em>normal distribution</em> 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 <em>z</em> follows a <em>standard normal distribution</em>, 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 <em>standard deviation</em> 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 <em>cumulative standard normal table</em> (or use some statistic program) to find the cumulative probability for <em>z</em> (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) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

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3 years ago