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3 years ago

The net of an isosceles triangular prism is shown. What is the surface area, in square units, of the triangular prism.

Mathematics

2 answers:

lions [1.4K]3 years ago

8 0

Answer:

I don't see the triangular prism.

Step-by-step explanation:

Inessa [10]3 years ago

6 0

here is the problem if you want to help them

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Can someone help me with this. Will Mark brainliest.

const2013 [10]

Answer:

The lines are perpendicular.

Step-by-step explanation:

If the lines are parallel, then the slopes are equal

If the lines are perpendicular, then the slopes are negative reciprocals

If neither of the above occur, then the lines are neither parallel or perpendicular.

y = 2x + 3 is in slope-intercept form, so the slope = 2

2y + x = 6

Solve for y: 2y = -x + 6

y = -1/2(x) + 3 so the slope = -1/2

2 and -1/2 are negative reciprocals, so the lines are perpendicular.

Note: Negative reciprocals have a product of -1. 2(-1/2) = -1

4 0

3 years ago

A recent college graduate is in the process of deciding which one of three graduate schools he should apply to. He decides to ju

romanna [79]

Answer:

For this case after conduct the ANOVA procedure they got an statistic of:

$F = 8.61$

With a p value of :

$p_v =P(F_{2,15} > 8.61) = 0.003$

Since the p value is lower than the significance level $\alpha=0.1$

We can reject the null hypothesis that the means are equal at the significance level provided.

$t = \frac{\bar X_i -\bar X_j}{s_p \sqrt{\frac{1}{n_i} +\frac{1}{n_j}}}$

For school 1 and 2 we have:

$t = \frac{646.7 -606.7}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 1.318$

For school 1 and 3

$t = \frac{646.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 4.068$

For school 2 and 3 we have:

$t = \frac{606.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 2.749$

So as we can see we have significant difference between the means of school 1 and 3, and school 2 and 3

Step-by-step explanation:

Previous concetps

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"

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{1}=\mu_{2}=\mu_{3}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=1,2,3$

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