Solve –3x = 36 for x. A. x = 9 B. x = 12 C. x = –12 D. x =

The means and mean absolute deviations of Sidney’s and Phil’s grades are shown in the table below which expression represents th

Shalnov [3]

The ratio of the difference of the two means to Sidney’s mean absolute deviation is; 4/3.28

<h3>How to find the Mean Absolute Deviation?</h3>

From the given table, we see that;

Mean grade of Sidney = 82

Mean grade of Phil = 78.

Mean absolute deviation of Sidney = 3.28

Mean absolute deviation of Phil = 3.96.

The difference between the two means of Sidney and Phil = 82 - 78 = 4.

Thus, the ratio of the difference of the two means to Sidney’s mean absolute deviation is; 4/3.28

Complete Question is;

The means and mean absolute deviations of Sidney’s and Phil’s grades are shown in the table below. Means and Mean Absolute Deviations of Sidney’s and Phil’s Grades Sidney Phil Mean 82 78 Mean Absolute Deviation 3.28 3.96 Which expression represents the ratio of the difference of the two means to Sidney’s mean absolute deviation?

Read more about Mean Absolute Deviation at; brainly.com/question/447169

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6 0

2 years ago

Hey! I need loads of help on this! If you could help it would be such a relief!

il63 [147K]

Answer 1: For the Parallelogram inside the rectangle the area = 48

Rectangle = 275 275+48 = 323

Answer 2: For the answer for the square inside the parallelogram = 36

The answer to the parallelogram = 120 120+36 = 156

Step-by-step explanation: #1: A = bh = 12 · 4 = 48

A = wl = 11 · 25 = 275

#2: A = a2 = 62 = 36

A = bh = 15 · 8 = 120

Sorry if u didn't have to add im not to good at this ;(

Hope that helped though!

4 0

3 years ago

The velocity of a car over five hours is given by v(t)=60ln(t+1),0≤t≤5 in kilometers per hour. What is the total distance travel

algol [13]

Step-by-step explanation:

as the velocity is not constant over time, we actually have to integrate v(t) over the interval 0<=t<=5 to get the distance.

v(t) = 60×ln(t + 1)

V(t) = 60 × integral(ln(t + 1)) between 0 and 5.

integral(ln(t + 1)) = (t + 1)ln(t + 1) - t + C

V(t) = 60 × ((t + 1)ln(t + 1) - t + C)

the distance traveled between t = 0 and t = 5 is then

60 × ((5 + 1)ln(5 + 1) - 5 + C) - 60 × ((0 + 1)ln(0 + 1) - 0 + C) =

= 60×(6ln(6) - 5 + C) - 60×(1ln(1) + C) =

= 60×(5.750556815... + C) - 60×C =

= 60×5.750556815... = 345.0334089... ≈ 345 km

7 0

2 years ago

Paired t‐Test for Mean Comparison with Dependent Samples To study the effects of an advertising campaign at a supply chain, seve

Lapatulllka [165]

Answer:

a) $t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.7 -0}{\frac{0.678}{\sqrt{5}}}=2.308$

$p_v =P(t_{(4)}>2.308) =0.0411$

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: $\mu_2- \mu_1 \leq 0$