If cot = 1.5 find tan

We have five samples of data: sample A with 30 successes of 50 cases, sample B with 600 successes of 1000 cases, sample C with 3

faust18 [17]

Answer:

C. with 3000 successes of 5000 cases sample

Step-by-step explanation:

Given that we need to test if the proportion of success is greater than 0.5.

From the given options, we can see that they all have the same proportion which equals to;

Proportion p = 30/50 = 600/1000 = 0.6

p = 0.6

But we can notice that the number of samples in each case is different.

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size

po = Null hypothesized value

p^ = Observed proportion

Since all other variables are the same for all the cases except sample size, from the formula for the test statistics we can see that the higher the value of sample size (n) the higher the test statistics (z) and the highest z gives the strongest evidence for the alternative hypothesis. So the option with the highest sample size gives the strongest evidence for the alternative hypothesis.

Therefore, option C with sample size 5000 and proportion 0.6 has the highest sample size. Hence, option C gives the strongest evidence for the alternative hypothesis

3 0

3 years ago

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$

Part b.

Consider V(3X + 2Y)

$\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}$

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

$3 X+2 Y \sim N(18,77)$

Thus,

$P(3 X+2 Y < 18)=0.5$

Part d.

Consider P(3X + 2Y < 28)

$Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}$

$\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}$

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

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The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0

2 years ago

SOMEONE HELP ME PLEASE

marin [14]

Answer:

Step-by-step explanation:

Direct variation problems can easily be solved with proportions, namely:

$\frac{x}{y} :\frac{2}{5}=\frac{4}{y}$ and cross multiply to get

2y = 20 so

y = 10

5 0

3 years ago

Read 2 more answers

It is difficult to manage data. For example, it is common for customers to move and for companies to go out of business. This is

Fantom [35]

Answer:

a) Data degradation

Step-by-step explanation:

These are the options for the question.

a) Data degradation

b) Data rot

c) Data security

d) Scattered data

From the question, we are informed about how difficult it's to manage data. For example, it is common for customers to move and for companies to go out of business. This case is an example of Data degradation. Data degradation can be regarded as

corruption of computer data, or gradual alterations of data, and this could be as a result of accumulation of failures in storage devices that store the data, this failure can be critical one or non-critical one

5 0

3 years ago

Eber ran 3 miles. He stopped to take a break every 34 mile. How many times did Eber stop, including his final stop at the end of

Elodia [21]

144 because when you divide them and simplify

3 0

2 years ago