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2 years ago

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A math teacher gave her class two tests. Twenty-five percent of the class passed both tests and 42 percent of the class passed o

nly the first test. What percent of the students who passed the first test also passed the second?

1 answer:

Norma-Jean [14]2 years ago

4 0

25% of the class who passed the first test also passed the second

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Show me how to find the answer to 6x1/3

mamaluj [8]

Answer:

2

Step-by-step explanation:

Simplify. Remember to follow PEMDAS and the left-> right rule.

In this case, multiply the whole number (6/1) with the numerator (1)

6 x 1 = 6

Next, simplify by dividing 6 with 3:

6/3 = 2

2 is your answer.

~

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3 years ago

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A rectangular red sticker is 8 millimeters wide and 5 millimeters tall. What is it’s area

PolarNik [594]

Answer:

40 milimeters

Step-by-step explanation:

8*5=40

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3 years ago

A rectangle prism with a length of 5 inches, a width of 4 inches, and a height of 3 inches.

Ivenika [448]

Answer:

▶◀

Step-by-step explanation:

n/a bc you didn't explain wut to do

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2 years ago

a password to a computer consists of 6 characters: a digit, a letter, a digit, a letter, a digit, and a letter in that order, wh

MrRa [10]

As per the concept of probability, there are 4,717,440 number of passwords are possible.

Probability:

In statistics, probability refers the favorable outcome of the particular event.

Given,

A password to a computer consists of 6 characters: a digit, a letter, a digit, a letter, a digit, and a letter in that order, where the numbers from 1 through 9 are allowed for digits.

Here we need to find how many different passwords are possible.

Total number of characters = 6

According to this, each password would have

5 digits of 10 digits: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], and

one of 26 letters: [A to Z].

Here there is no repeats are allowed, and assuming only upper case letters are valid and the letter is the last character, then we can form 10*9*8*7*6 * 26 = 786,240 such passwords.

If the repetition is allowed then number of possible passwords is

786,240 * 6 = 4,717,440.

To know more about Probability here.

brainly.com/question/11234923

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11 months ago

A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Const

Ludmilka [50]

Answer:

$0.609 \leq \sigma \leq 0.772$

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation

s=0.68 represent the sample standard deviation

$\bar x =98.90$ represent the sample mean

n=98 the sample size

Confidence=90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=98-1=97$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:

$\chi^2_{\alpha/2}=120.990$

$\chi^2_{1- \alpha/2}=75.282$

And replacing into the formula for the interval we got:

$\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}$

$0.371 \leq \sigma^2 \leq 0.596$

Now we just take square root on both sides of the interval and we got:

$0.609 \leq \sigma \leq 0.772$

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

3 0

3 years ago