Please help. it’s 7th grade math. (mathematics)))

A line segment is sometimes/always/never similar to another line segment, because we can sometimes/never/always map one into the

Fantom [35]

Answer:

A line segment is always similar to another line segment, because we can always map one into the other using only dilation a and rigid transformations

Step-by-step explanation:

we know that

A dilation is a Non-Rigid Transformations that change the structure of our original object. For example, it can make our object bigger or smaller using scaling.

The dilation produce similar figures

In this case, it would be lengthening or shortening a line. We can dilate any line to get it to any desired length we want.

A rigid transformation, is a transformation that preserves distance and angles, it does not change the size or shape of the figure. Reflections, translations, rotations, and combinations of these three transformations are rigid transformations.

so

If we have two line segments XY and WZ, then it is possible to use dilation and rigid transformations to map line segment XY to line segment WZ.

The first segment XY would map to the second segment WZ

therefore

A line segment is always similar to another line segment, because we can always map one into the other using only dilation a and rigid transformations

6 0

3 years ago

Help I don’t get it at all

Naily [24]

The answer to your question is n is equal to 70

5 0

3 years ago

Read 2 more answers

Ch.3 Test Savings Accounts

Harlamova29_29 [7]

Step-by-step explanation:

5000 x 3 + 3.5% of 5000 x 3 = the answer

8 0

3 years ago

X intercept Of v^2 +5v+6=0

postnew [5]

Answer:

<h2>x-intercepts:</h2><h2>x = -2 and x = -3 ⇒ (-2, 0) and (-3, 0).</h2>

Step-by-step explanation:

$f(v)=v^2+5v+6\\\\\text{The x-intercept is for y = 0.}\\\\v^2+5v+6=0\\\\v^2+2v+3v+6=0\\\\v(v+2)+3(v+2)=0\\\\(v+2)(v+3)=0\iff v+2=0\ \vee\ v+3=0\\\\v+2=0\qquad\text{subtract 2 from both sides}\\v=-2\\\\v+3=0\qquad\text{subtract 3 from both sides}\\v=-3$

3 0

3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago