All categories

3 years ago

How many quarts and cups are in 2 gallons

Mathematics

2 answers:

MrMuchimi3 years ago

7 0

There are 8.25 cups and quarts in two gallons

Len [333]3 years ago

3 0

There are 8 quarts in 2 gallons.

There are 4 cups in a quart.

There are 32 cups in 2 gallons.

Hope this helps! :)

You might be interested in

Please help me!!! :(

eduard

Answer:

1. ecologist

2.species

3. population

4.habitat

Step-by-step explanation:

5 0

3 years ago

Find the focus of the parabola whose equation is y=1/8 x^2

Mama L [17]

Answer:

The focus is: $(0,2)$

Step-by-step explanation:

Given

$y = \frac{1}{8}x^2$

Required

Determine the focus

The focus of a parabola

$(x- h)^2 = 4p(y - k)$

is:

$(h,k+p)$

So, we have:

$y = \frac{1}{8}x^2$

Cross multiply

$8y = x^2$

Rewrite as:

$x^2 = 8y$

Rewrite as:

$(x - 0)^2 = 8(y - 0)$

Express 8 as 4 * 2

$(x - 0)^2 = 4 * 2(y - 0)$

By comparison with: $(x- h)^2 = 4p(y - k)$

$h = 0$ $p =2$ $k = 0$

So, the focus $(h,k+p)$ is:

$(h,k+p) = (0,0+2)$

$(h,k+p) = (0,2)$

7 0

3 years ago

The sum of the angle measures of a quadrilateral is 360°. Write and solve an equation to find the value of X.

uranmaximum [27]

Answer:

1). x° + 100° + 120° + 100° = 360°

2). x = 40

Step-by-step explanation:

Formula for the sum of the interior angles of a polygon = $(n-2)\times 180$

Where n = number of sides of the polygon

From the picture attached,

For n = 4,

Sum of the interior angles of the given polygon = (4 - 2) × 180°

= 360°

Therefore, equation for the sum of interior angles will be,

x° + 100° + 120° + 100° = 360°

x° + 320° = 360°

x° = 360° - 320°

x° = 40°

6 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

Which of the following is the equation of a line that passes through the point (3,2) and is parallel to the y-axis

liubo4ka [24]

X= 3
hope this helps!

8 0

4 years ago