16 1/9+ -4 2/3 help please??

Answers

Answer 1
Answer:

Answer:

11 4/9

Step-by-step explanation:

ANDDD the answer is


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Which fraction is larger: -3/4 or –2/4

Answers

Answer:3/4

Step-by-step explanation: If you weigh a brick at 2/4 lbs, it will weigh less than a brick that weighs 3/4

Please answer asap.........

Answers

Answer:

1.4 i would wait for someone else to answer it I'm not sure

Step-by-step explanation:

Write in slope intercept form an equation of the line that passes through the given points. (0,4) (4,8) and (-2,3) (-4,4)

Answers

Answer:

y-8=1(x-4)

Step-by-step explanation:

I used the equations y-y/x-x and y-__y__=__m__(x-__x__)

y=-8=1(x-4) hope that helped!

Paired t‐Test for Mean Comparison with Dependent Samples To study the effects of an advertising campaign at a supply chain, several stores are randomly selected with the following observed before‐ and after‐advertising monthly sales revenues: Store number 1 2 3 4 5
Old sales revenue (mil. $) 5.2 6.5 7.2 5.7 7.6
New sales revenue (mil. $) 6.4 7.8 6.8 6.5 8.2
Let μ₁ and μ₂ be the means of old and new sales revenues, both in millions of dollars per month.
(a) At α = 0.05, test H0: μ2 ≤ μ1 versus H1: μ2 > μ1. Sketch the test. Interpret your result.
(b)Sketch and find the p‐value of the test. Would you reject H0 if α = 0.01?

Answers

Answer:

a) t=(\bar d -0)/((s_d)/(√(n)))=(0.7 -0)/((0.678)/(√(5)))=2.308  

p_v =P(t_((4))>2.308) =0.0411

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

b) The p value is illustrated on the figure attached.

If we select \alpha=0.01 we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

1=test value old , 2 = test value new

1: 5.2 6.5 7.2 5.7 7.6

2: 6.4 7.8 6.8 6.5 8.2

The system of hypothesis for this case are:

Null hypothesis: \mu_2- \mu_1 \leq 0

Alternative hypothesis: \mu_2 -\mu_1 >0

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: 1.2, 1.3, -0.4, 0.8, 0.6

The second step is calculate the mean difference  

\bar d= (\sum_(i=1)^n d_i)/(n)=0.7

The third step would be calculate the standard deviation for the differences, and we got:

s_d =(\sum_(i=1)^n (d_i -\bar d)^2)/(n-1) =0.678

The 4 step is calculate the statistic given by :

t=(\bar d -0)/((s_d)/(√(n)))=(0.7 -0)/((0.678)/(√(5)))=2.308

The next step is calculate the degrees of freedom given by:

df=n-1=5-1=4

Now we can calculate the p value, since we have a right tailed test the p value is given by:

p_v =P(t_((4))>2.308) =0.0411

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.  

Part b

The p value is illustrated on the figure attached.

If we select \alpha=0.01 we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis.

Please help me im stuck

Answers

Answer:

A. by adding an identical classroom and allowing students t use electric ans in it

Step-by-step explanation:

Because Fans are opposite to natural Air

1. Suppose you have a fair 6-sided die with the numbers 1 through 6 on the sides and a fair 5-sided die with the numbers 1 through 5 on the sides. What is the probability that a roll of the six-sided die will produce a value larger than the roll of the five-sided die? 2. What is the expected number of rolls until a fair five-sided die rolls a 3?

Answers

Answer:

a. 0.5 or 50%

b. 5 rolls.

Step-by-step explanation:

a. There are 30 possible outcomes for this experiment, the sample space for the outcomes in which the six-sided die produces a value larger than the roll of the five-sided die is:

S={6,1; 6,2; 6,3; 6,4; 6,5; 5,1; 5,2; 5,3; 5,4; 4,1; 4,2; 4,3; 3,2; 3,1; 2,1}

There are five outcomes when rolling a 6, four when rolling a 5, three when rolling a 4, two when rolling a 3 and one when rolling a two.

The probability is:

P = (5+4+3+2+1)/(5*6)=0.5

b. The probability of rolling a 3 on the five-sided die is 1 in 5 or 0.20. The expected number of rolls until a fair five-sided die rolls a 3 is:

E(x=1) = (1)/(p(x))=(1)/(0.2)= 5\ rolls