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3 years ago

Which statement best compares a line and a point

1 answer:

Volgvan3 years ago

6 0

A line has 0 endpoints and extends forever in 2 directions and a point has 1 endpoint and extends forever in 0 directions.

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The product of 6g2 and -2g3 is equals to

ryzh [129]

Answer:

We conclude that:

$\left(6g^2\:\right)\times \left(-2g^3\right)=-12g^5$

Step-by-step explanation:

Given

6g²

-2g³

The product of 6g² and -2g³ can be determined by multiplying each other such as:

$(6g^2\:)\times \:(-2g^3)$

Apply the rule: a(-b) = -ab

$\left(6g^2\:\right)\times \left(-2g^3\right)=-6g^2\times \:\:2g^3$

Apply exponent rule: $a^b\times \:a^c=a^{b+c}$

$=-6g^{2+3}\times \:2$

$=-6g^5\times \:2$

$=-12g^5$

Therefore, we conclude that:

$\left(6g^2\:\right)\times \left(-2g^3\right)=-12g^5$

8 0

3 years ago

Please please help. Thank you

Diano4ka-milaya [45]

Answer is x^3(x+3)(x+1)

5 0

3 years ago

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores

EastWind [94]

Answer:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

Step-by-step explanation:

We have the following info given from the problem

$\bar X_1 = 52$ sample mean for this year

$s_1= 13$ sample deviation for this year

$n_1 = 75$ random sample selected for this year

$\bar X_2 = 49$ sample mean for last year

$s_2= 11$ sample deviation for last year

$n_1 = 53$ random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df= n_1 +n_2 -2 = 75+53-2= 126$

The confidence level is 0.95 and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ so then the critical value for this case is :

$t_{\alpha/2}= 1.979$

The margin of error would be:

$ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300$

And the confidence interval would be given by:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

6 0

3 years ago

The midpoint of the sides of triangle ABC are labeled D, E, and F. If the perimeter of triangle DEF is 24 find the perimeter of

AleksAgata [21]

Ok well the only thing that makes since to me is that the triangle is a Equilateral Triangle, so all the sides are going to be the same. And if the perimeter is 24, then you divide that by 3, and you get 8. And if DEF is the midpoints of triangle ABC, then DEF is an upside down triangle in the triangle ABC. So all the sides of ABC are double whatever DEFs are. And DEFs were 8, so 8 times 2 is 16. And to get the perimeter you add that together 3 times and you get 48. So i'm not sure exactly how you would do this, but i'm pretty sure this is the way. If not i'm sorry i couldn't help. :(

8 0

3 years ago

Read 2 more answers

Write the interval (5,100]as an inequality and using set notation

sveticcg [70]

5<x<=100 x∈(5;100] hope this helps

4 0

2 years ago