What is the value ofcos pie value

Answers

Answer 1
Answer: cos(pi) = -1

You should have this memorized along with other values such as
sin(0) = 0
cos(0) = 1
sin(pi/2) = 1
cos(pi/2) = 0
sin(pi) = 0
cos(pi) = -1
sin(3pi/2) = -1
cos(3pi/2) = 0

and remember that tan = sin/cos


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(-3l^2w^3)(2lw^4) simplify express using exponents.

Answers

The simplified algebraic expression using exponents is (-3l^2w^3)(2lw^4) simplifies to -6l^3w^7.

To simplify the given expression using exponents, follow these steps:

Multiply Coefficients: Multiply the coefficients (-3) and (2) to get -6.

Combine Like Bases: For the variables with the same base (l and w), add the exponents when they are multiplied together.

Here, l^2 * l^1 = l^(2+1) = l^3, and w^3 * w^4 = w^(3+4) = w^7.

Final Simplified Expression: Combine the results from steps 1 and 2 to get -6l^3w^7.

Therefore, the simplified expression using exponents is (-3l^2w^3)(2lw^4) simplifies to -6l^3w^7.The expression has been simplified using the rules of exponentiation. This simplification helps in reducing the complexity of the expression and making calculations easier.

Learn more about algebraic expression here:

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Answer:

Step-by-step explanation:

(-3)(2)= -6

(l^2w^3)(lw^4) = l^3w^7

-6l^3w^7

Let X be a binomially distributed random variable with parameters n=500 and p=0.3. The probability that X is no larger than one standard deviation above its mean is closest to which of the following? a. 0.579
b. 0.869
c. 0.847
d. 0.680

Answers

Answer:

c. 0.847

Step-by-step explanation:

From the given information;

X \sim Binomial (500,0.3)

Given that n = 500 which is too large, binomial distribution can now be approximated to

N ( \mu , \sigma^2)

where;

\mu = np

\mu =500 * 0.3

\mu =150

\sigma^2= np(1-p)

\sigma^2= 500 * 0.3(1-0.3)

\sigma^2= 150(0.7)

\sigma^2= 105\n

P(X \leq \mu + \sigma ) = ( (X-\mu)/(\sigma) \leq 1)

P(X \leq \mu + \sigma ) = ( Z \leq 1)

P(X \leq \mu + \sigma ) =\Phi (1)

From the z table

P(X \leq \mu + \sigma ) =0.841

Thus, our value is closest to the option c which 0.847

People tend to evaluate the quality of their lives relative to others around them (Frieswijk et al., 2004). In one study, researchers conducted interviews with n = 9 frail elderly people. During the interview, each person was compared with a fictitious person who was worse off than the elderly person. The scores below are the measures from a life-satisfaction scale for the elderly sample. Assume that the average score on this scale in the population is u = 20. Are the data sufficient to conclude that the elderly people in this sample are either significantly more or less satisfied than others in the general population? The life-satisfaction scores for the sample are 18, 23, 24, 22, 19, 27, 23, 26, 25. a. Which kind of t-test should you use? b. How many tails should the test have? Circle a word or phrase in the problem that told you this. c. State the null and alternative hypotheses in statistical notation: d. Determine the critical t using an alpha = .05. Sketch the null distribution, note the location of the critical t, and shade the critical region. e. Calculate the t-statistic and plot it on the sketch you drew above. f. Make a decision (either reject the null or fail to reject it)

Answers

Answer:

Step-by-step explanation:

Hello!

Research.

n=9 frail elderly were interview and compared to a fictitious person who was worse off then the interviewee, a life-satisfaction score was determined for each person.

18, 23, 24, 22, 19, 27, 23, 26, 25

Assuming that the population average score is μ= 20, the researchers think that the elderly in the sample are more or less satisfied than others in the general population.

a. You have the information of one sample, assuming this sample has a normal distribution and each elderly interviewed is independent, then the t-test of choice is a one-sample t-test.

b. and c. If you say that the elderly are "more or less" satisfied than the others, this means that they are either as satisfied as to the general population or not satisfied as to the general population. Symbolically:

H₀: μ = 20

H₁: μ ≠ 20

This is a two-tailed test, meaning, you will have two critical regions.

d.  

α: 0.05

Left critical value: t_(n-1;/\alpha 2) = t_(8; 0.025)= -2.306

Right critical value: t_(n-1;1-\alpha /2) = t_(8;0.975) = 2.306

e.

t_(H_0)= (X[bar]-Mu)/((S)/(√(n) ) ) ~t_(n-1)

X[bar]= 23

S= 3

t_(H_0)= (23-20)/((3)/(√(9) ) )= 3

f.

Considering that the calculated t-value is greater than the right critical value, the decision is to reject the null hypothesis, so using a significance level of 5% you can conclude that the average life-satisfaction score of the elderly is different than 20.

I hope it helps!

See images for the following problem

Answers

I can’t see it that well sorry

PLEASE HELP The probability distribution for a
random variable x is given in the table.

Answers

1

Step-by-step explanation:

p(x≤20)=p(x=-10) +p(x=-5)+p(X=0) +p(x=5) +p(x=10)+ p(X=15)+p(X=20)

This, p(X≤20)=0.20+0.15+0.05+0.1+0.25+0.1+0.15

=1

Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is the temperature in Celsius at a height of kilometers above the planet's surface. The relationship is as follows.

Answers

The question is incomplete, the complete question is below:

Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is the temperature in Celsius at a height of kilometers above the planet's surface. The relationship is as follows: T(h) = 30.5 -2.5h.

a) Calculate T^-1(x)

b) T^-1(15)

Answer:

a) T^(-1) (x)= (30.5-x)/(2.5)

b) 6.2 °C

Step-by-step explanation:

a) T(h) = 30.5 - 2.5h\n2.5h = 30.5-T(h)\nh = (30.5-T(h))/(2.5)\n T^(-1) (x)= (30.5-x)/(2.5)

b) T^(-1) (x)= (30.5-x)/(2.5)\n\nT^(-1) (15)= (30.5-15)/(2.5)=6.2