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4 years ago

How do you find the number of possible choices for a 2 digit password that is greater than 19?

1 answer:

3 0

There probably is a formula to do this, but a simple way to find this one is to just think of all two digit numbers. That is 00-99. All you do is count the number of possible choices from 20-99. So there are 80 possible choices.

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Analogies: Write the letter indicating the pair of words that best expresses a relationship[ similar to that of the original pai

allochka39001 [22]

The answer to this is d

8 0

3 years ago

V + 9 over 3= 8<br> what is v?

stira [4]

V= 15 steps are down below

6 0

3 years ago

Read 2 more answers

Some conference-goers saunter over to the Healthy Snack Box Machine, where they each choose one of ﬁve kinds of fruit, one of th

nalin [4]

Answer: There are 90 snack boxes.

Step-by-step explanation:

Given : The number of kinds of fruits = 5

The number of kinds of herbal teas = 3

The number of kinds of flavors of wrap sandwich = 6

Then by using the fundamental principal of counting, the number of possible snack are there will be :_

$5\times3\times6= 90$

Therefore, the number of possible snacks = 90

4 0

3 years ago

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

melamori03 [73]

Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{236100}{8} = 29512.5$

Sum of squares of differences = 3408750

$S.D = \sqrt{\frac{3408750}{7}} = 697.82$

Population mean, μ = $30,000

Sample mean, $\bar{x}$ = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars}$ We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975$

Now,

$t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

8 0

4 years ago

Which 3 values would make this inequality true? 48 < 6n

tekilochka [14]

Answer:

The three values are: 9, 10 and 11

5 0

3 years ago