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Aleonysh [2.5K]

3 years ago

5

Blake looks at his checking account balance regularly. His three lowest balance during the year are as given. - $20 < -$12&lt

; - $ 3 select from the drop-down menus to correctly complete the statement The inequality statement shows that - $20 is a (lower or higher) which one is it Balance than -$12 and - $12 is a (Lower or higher) Which one is it balance than - $3

2 answers:

Naya [18.7K]3 years ago

5 0

I believe the answers are lower and lower. -20 is lower than -12, and -12 is lower than -3

Montano1993 [528]3 years ago

4 0

hey that is my name Blake -20 lower -12 high and -3 higher

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What number is equivalent to 15 and 25

Gwar [14]

Answer:

yo are you talking about like absolute value? then it's | -15 | and | -25 |

Step-by-step explanation:

3 0

3 years ago

Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random

Katyanochek1 [597]

Answer:

a) Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

The mean squares between groups are given by:

$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

$MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544$

And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

And the p value is given by:

$p_v= P(F_{2,15} >11.326) = 0.00105$

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

b) $(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}$

The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part a

Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

If we assume that we have $3$ groups and on each group from $j=1,\dots,6$ we have $6$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

The mean squares between groups are given by:

$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

$MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544$

And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

And the p value is given by:

$p_v= P(F_{2,15} >11.326) = 0.00105$

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

Part b

For this case the confidence interval for the difference woud be given by:

$(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}$

The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

7 0

3 years ago

A. 42<br> b. 27<br> c. 12<br> d. 18

hodyreva [135]

The answer is B simple

7 0

2 years ago

Find The volume of a cilinder

Alinara [238K]

Answer:

60

Step-by-step explanation:

calculated it with scientific calculator

7 0

3 years ago

PLEASE HELP PLEASE!!!

defon

Answer:

We conclude that at x = 0 and x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

Therefore, the solution to f(x) = g(x) is:

x = -1

x - 0

Step-by-step explanation:

Given the table

x f(x) = 2ˣ - 1 g(x) = 1/2x

-2 -3/4 -1

-1 -1/2 -1/2

0 0 0

1 1 1/2

2 3 1

If we carefully observe, we can determine that

at x = 0, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

In other words,

at x = 0

f(0) = 0

g(0) = 0

Thus,

at x = 0

f(x) = g(x)

Also at x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

In other words,

at x = -1

f(-1) = -1/2

g(-1) = -1/2

Thus,

at x = -1

f(x) = g(x)

Summary:

Thus, we conclude that at x = 0 and x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

Therefore, the solution to f(x) = g(x) is:

x = -1

x - 0

6 0

2 years ago