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3 years ago

Which shows 2-6 ///////////////////////////////////////////////////////

Mathematics

2 answers:

Leokris [45]3 years ago

7 0

Answer:

Step-by-step explanation:

(1/2⁶)

IgorLugansk [536]3 years ago

3 0

Answer:

C

Step by step explanation:

(1/2^6)

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HELLPPPPPP<br> PLZ <br> Thank you

uranmaximum [27]

Answer:

it is 25 words per minuet

Step-by-step explanation:

7 0

3 years ago

15c² - 5c. Factorise completely

KATRIN_1 [288]

Looking at the problem, what we must do to complete this question is to completely factor the expression that was provided. The expression that was provided is $15c^2 - 5c$ .

The first step that we must do is to take a look at the expression and see what the two pieces of the expression have in common. We can see that both $15c^2$ and $-5c$ have the number 5 and the variable c associated with them so we can factor out those two.

<u>Factor out 5c</u>

$15c^2 - 5c$

$5c(\frac{15c^2}{5c}) - 5c(\frac{5c}{5c})$

$5c(3c) - 5c(1)$

$5c(3c - 1)$

Now we have completely factored out the expression that was provided in the problem statement and we came to final answer of $5c(3c - 1)$ .

7 0

2 years ago

Read 2 more answers

A new anxiety medication has been manufactured and a study is being conducted to determine whether its effectiveness depends on

Irina18 [472]

The test statistics is used to determine if the anxiety medication can happen under a null hypothesis

The test statistics that is an appropriate hypothesis test is $z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}$

<h3>How to determine the test statistic</h3>

<u>40 milligrams of medication</u>

Patients = 60

Lower stress level patients = 21

The mean of this medication is:

$\bar x = \frac{x}{n}$

So, we have:

$\bar x_1 = \frac{21}{60}$

$\bar x_1 = 0.35$

<u>75 milligrams of medication</u>

Patients = 85

Lower stress level patients = 34

The mean of this medication is:

$\bar x = \frac{x}{n}$

So, we have:

$\bar x_2 = \frac{34}{85}$

$\bar x_2 =0.4$

The test statistic is then calculated as:

$z =\frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\bar x_1(1 - \bar x_1)}{n_1} + \frac{\bar x_2(1 - \bar x_2)}{n_2}}}$

The equation becomes

$z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}$

Hence, the test statistics that is an appropriate hypothesis test is $z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}$