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3 years ago

6

A researcher believes that headaches can be relieved by drinking water flavored by

1 answer:

olganol [36]3 years ago

4 0

Answer: A. Experimental Study

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PLEASE HELP ME WITH THIS QUESTION ^^

AURORKA [14]

I’m sorry I don’t get that whole situation

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2 years ago

A forest fire covers 2006 acres at time t equals 0. the fire is growing at a rate of 8 startroot t endroot acres per hour, where

MissTica

<span>Answer: We'll have to integrate the growth rate with respect to time, without forgetting the amount of area covered at the start. A = 2010 + â« 8t1/2 dt (as t goes from 0 to 24) A = 2010 + [(16/3) t3/2] (as t goes from 0 to 24) A = 2010 + 627 = 2637 acres We don't integrate the 2010 here because whatever effects of the passing of time have already taken place over that area. We just treat it as a constant and add it to the result of integration.</span>

8 0

3 years ago

I need help with 29, 32, 35, 38, 43, 46, 49, 44, 47, 50. ASAP

gogolik [260]

29. d=-4
32. h=45
35. n=76
38. t=5
43. x=-11
46. x≈0.7
47. x≈1.6
49. d=-2
50. t=1/2

7 0

3 years ago

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

3 0

2 years ago

You are in a bike race. When you get to the first checkpoint, you are 2525 of the distance to the second checkpoint. When you ge

svetoff [14.1K]

1/10
or maybe even 16?
not too sure though

3 0

3 years ago