What is the exact value of sin 30° ?

HELP ME !!!!!!! Which of the following best describes a single fixed point that is the same

s344n2d4d5 [400]

Answer:

C. Focus

.............

8 0

2 years ago

Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance

lana66690 [7]

Answer:

Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20% of users develop nausea.

Let p = population proportion of users who develop nausea

So, Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Here, null hypothesis states that 20% of users develop nausea.

And alternate hypothesis states that more than 20% of users develop nausea.

The test statistics that would be used here is One-sample z proportion test statistics.

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of users who develop nausea in a sample of 241 subjects = $\frac{54}{241}$

n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

6 0

3 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

i)

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago

if a pharmacy technician makes as error resulting in serious consequences who can be held liable in court for error of negligenc

beks73 [17]

The very pharmacy technician must be held guilty

3 0

3 years ago

Simplify:

OverLord2011 [107]

Step-by-step explanation:

Simplify :

x4−6x3+2x−7+7x3−x+5x2+2−x4

Rearranging and collecting the like terms, we get:

=(x4−x4)+(−6x3+7x3)+5x2+(2x−x)+(−7+2)

=(1−1)x4+(−6+7)x3+5x2+(2−1)x+(−7+2)

=(1)x3+5x2+(1)x−7+2

=x3+5x2+x−5

8 0

3 years ago