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cricket20 [7]
3 years ago
5

In the first quadrant you start at 2,4 and move 5 units right

Mathematics
1 answer:
NikAS [45]3 years ago
5 0

Answer:

Thanks for asking!

Step-by-step explanation:

(2,9)

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What is the slope of y = 2 + 5?
brilliants [131]

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Using the slope-intercept form, the slope is 0 .

Step-by-step explanation:

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Unit 5. 1) Please help. What is the volume of the cone?
SVEN [57.7K]

Answer:

I think the correct answer is 27 so option c. :)

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3 years ago
A. Use the​ one-mean t-interval procedure with the sample​ mean, sample​ size, sample standard​ deviation, and confidence level
PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668

Margin of error, E = t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

M.E = \frac{22.773 - 17.227}{2} = 2.773

c)M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}

= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

4 0
3 years ago
21.56 S-AQ) Healthy women aged 18 to 40 participated in a study of eating habits. Subjects were given bags of potato chips and b
IRISSAK [1]

Answer:

There is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Step-by-step explanation:

The question is incomplete as some information is not provided, please refer below the remaining information of the question.

Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation):

Group          n     bar{x}         s

Unrestrained  9     63.83      24.72

Restrained  11     34      39.8

a. The standard error of the difference between sample means  ( ± 0.0001 )  is:

b. The critical value  ( ± 0.001 )  from the t distribution for confidence interval 80 % using the conservative degrees of freedom is:

c. Give a 80 % confidence interval   ( ± 0.01 )  that describes the effect of restraint:

Answer:

a) The standard error of the difference between sample means:

S E  =  √ s 2 1 /  n 1  +  s 2 2 /n 2

    =  √ s e 2 1 +  s e 2 2

    =  √ 24.72 2 +  39.8 2

    =  46.85

b) The degrees of freedom:

D f  =  n 1  + n 2  − 2

=  9  +  11  −  2

= 18

The confidence level  =  0.80

The significance level,  α  =  0.20

The critical value from the t distribution for confidence interval 80%:

t c r i t i c a l  =  t α / 2 ,  d f  =  t 0.10 , 18  =  ±  1.33

c) The 80% confidence interval:

( μ 1 − μ 2 ) = ( ¯ x 1 − ¯ x 2 )  ±  t ⋅ S E

= ( 63.83 − 34 ) ±  1.33 × 46.85  

= 29.83 ± 62.31

− 32.48 < ( μ 1 − μ 2 ) <  92.14

As the interval contains the zero. So, it can be concluded that there is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

5 0
4 years ago
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