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3 years ago

Subtraction with renaming estimate 2 1/6- 1 2/7

Mathematics

1 answer:

Snezhnost [94]3 years ago

3 0

13/6-9/7
Find the Least Common Denominator (LCD)
In this case, 42.
(7/7)(13/6)-(6/6)(9/7)
37/42 is your answer

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Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.2 significance level. The

sattari [20]

Answer:

C. H0 : p = 0.8 H 1 : p ≠ 0.8

The test is:_____.

c. two-tailed

The test statistic is:______p ± z (base alpha by 2) $\sqrt{\frac{pq}{n} }$

The p-value is:_____. 0.09887

Based on this we:_____.

B. Reject the null hypothesis.

Step-by-step explanation:

We formulate null and alternative hypotheses as proportion of people who own cats is significantly different than 80%.

H0 : p = 0.8 H 1 : p ≠ 0.8

The alternative hypothesis H1 is that the 80% of the proportion is different and null hypothesis is , it is same.

For a two tailed test for significance level = 0.2 we have critical value ± 1.28.

We have alpha equal to 0.2 for a two tailed test . We divided alpha with 2 to get the answer for a two tailed test. When divided by two it gives 0.1 and the corresponding value is ± 1.28

The test statistic is

p ± z (base alpha by 2) $\sqrt{\frac{pq}{n} }$

Where p = 0.8 , q = 1-p= 1-0.8= 0.2

n= 200

Putting the values

0.8 ± 1.28 $\sqrt{\frac{0.8*0.2}{200} }$

0.8 ± 0.03620

0.8362, 0.7638

As the calculated value of z lies within the critical region we reject the null hypothesis.

8 0

3 years ago

Explain how to estimate 368+231 two different ways

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

368..........400

231...........200

400+200 = 600

300+200 =500

68.......70

31........30

70+30 =100 500+100 =600

4 0

4 years ago

Which graph is the solution of the following system

scoray [572]

Answer:

It is A(maybe C but doubt it)

Step-by-step explanation:

5 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

HELP PLEASE HELP!!!!! 20 POINTS!!!

ivann1987 [24]

The answer should be C. By combining like terms you can find the answer. 1.75a+2.75a = 4.5a
2.25b + 1.75b = 4b
2.25c + 1.25c = 3.5c

3 0

4 years ago

Read 2 more answers