All categories

4 years ago

What is 10.417 rounded to the nearest hundredth?

Mathematics

1 answer:

Zarrin [17]4 years ago

7 0

10.417 rounded to the nearest hundredth is 10.42

Hope it helps!

You might be interested in

Help me with this please marking brainliest!

vodomira [7]

Answer:

A) ± $6\sqrt{2}$

Step-by-step explanation:

3 0

3 years ago

At the time that a 55 foot tall tree casts a shadow that is 32 feet long, a nearby woman is 5.5 feet tall. Which measure is the

Tasya [4]

Answer:

3.2ft long shadow

Step-by-step explanation:

In this problem we are given a ratio between the tree/shadow and woman/shadow therefore we are given 3 values and 1 unknown then we can use the Rule of Three to calculate the length of the shadow which we will refer to as x. The Rule of Three simply multiplies the diagonal values and divides by the third value in order to get the missing unknown like so...

55ft tall <======> 32ft long shadow

5.5ft tall <======> x ft long shadow

(5.5 * 32) / 55 = 3.2ft long shadow

Finally, we can see that the woman's shadow would be a 3.2ft long shadow

8 0

3 years ago

Plot J(-2,8), K(0,8), L(1,0), and M(1,-2) in a coordinate plane. Then determine whether JK and LM are congruent. A. Cannot be de

mr_godi [17]

Answer:

C. JK and LM are congruent

Step-by-step explanation:

Given the coordinates J(-2,8), K(0,8), L(1,0), and M(1,-2), in other to determine whether JK is congruent to LM, we need to determine if the length of side JK and LM are equal i.e JK = LM

To get JK:

We will find the distance between the coordinates J(-2,8), K(0,8) to get the length JK using the formula:

JK =√(x2-x1)²+(y2-y1)²

JK = √(0-(-2))²+(8-8)²

JK = √2²+0²

JK = √4

JK = 2

For the length of LM with coordinate L(1,0), and M(1,-2)

LM = √(1-1)²+(-2-0)²

LM = √0²+(-2)²

LM = √0²+2²

LM = √0+4

LM =√4

LM = 2

Since JK = LM = 2 units, this means that JK and LM are congruent.

7 0

3 years ago

HEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLP What is the simplified value of the expression below?

olganol [36]

Answer:

100

Step-by-step explanation:

start with what's in the parenthesis':

8 (9.75 - 3.25) + 12 x 4 = 8 (6.50) + 12 x 4

then take the number outside of the parenthesis' time the number inside of them:

8 (6.50) + 12 x 4 = 52 + 12 x 4

now, take the product of the two number on the right side of the addition sign:

52 + 12 x 4 = 52 + 48

finally, add the final two numbers together:

52 + 48 = 100

6 0

3 years ago

Read 2 more answers

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