What is the greatest common factor of 36, 34, and 22?

<span>Considering the visitor count is likely rounded to the nearest hundred thousand, the precise figure could range from 350,000 to 449,999. If rounded to the nearest ten thousand, it would be between 395,000 and 404,999.</span>

Answer:

a) Null hypothesis:

Alternative hypothesis:

b)

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c)

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d)

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

represents the average of the samples

denotes the standard deviation of the samples

indicates the number of samples

is the value we are examining

defines the significance level for the test.

t represents the specific statistic of interest

indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:

Alternative hypothesis:

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

(1)

Part b

In this scenario t=-2.3,

Initially, we need to calculate the degrees of freedom

Since this is a left-tailed test, the p-value is determined by:

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8,

We need to find the degrees of freedom

For the left-tailed test, the p-value is given by:

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6,

Firstly, we find the degrees of freedom

Since we are conducting a left-tailed test, the p-value is calculated as:

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

The total is 4 boxes, as you don't count the.6 since it's not a complete box.

Let x represent the number of caps

cost per cap = $6

cost for x caps = 6x

shipping charge = $25

overall budget = $1000

We can set up the following inequality:

Because the total amount cannot exceed 1000, we have

25 + 6x ≤ 1000

6x ≤ 975

x ≤ 162.5

rounding,

x ≤ 163

This means she can purchase a maximum of 163 caps.

Answer:

Step-by-step explanation:

Considering the equation

Sin(5x) = ½

5x = arcSin(½)

5x = 30°

Then,

The general formula for sin is

5θ = n180 + (-1)ⁿθ

Dividing throughout by 5

θ = n•36 + (-1)ⁿ30/5

θ = 36n + (-1)ⁿ6

The solution range is

0<θ<2π which means 0<θ<360

First solution

When n = 0

θ = 36n + (-1)ⁿθ

θ = 0×36 + (-1)^0×6

θ = 6°

When n = 1

θ = 36n + (-1)ⁿ6

θ = 36-6

θ = 30°

When n = 2

θ = 36n + (-1)ⁿ6

θ = 36×2 + 6

θ = 78°

When n =3

θ = 36n + (-1)ⁿ6

θ = 36×3 - 6

θ = 102°

When n=4

θ = 36n + (-1)ⁿ6

θ = 36×4 + 6

θ = 150

When n=5

θ = 36n + (-1)ⁿ6

θ = 36×5 - 6

θ = 174°

When n = 6

θ = 36n+ (-1)ⁿ6

θ = 36×6 + 6

θ = 222°

When n = 7

θ = 36n + (-1)ⁿ6

θ = 36×7 - 6

θ = 246°

When n =8

θ = 36n + (-1)ⁿ6

θ = 36×8 + 6

θ = 294°

When n =9

θ = 36n + (-1)ⁿ6

θ = 36×9 - 6

θ = 318°

When n =10

θ = 36n + (-1)ⁿ6

θ = 36×10 + 6

θ = 366°

When n = 10 surpasses the θ range

Thus, the solutions range from n =0 to n=9

Therefore, there are 10 solutions within the interval 0<θ<2π