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2 years ago

Points a b and c ate noncollinear so therefore they are noncoplanar

Mathematics

1 answer:

Kamila [148]2 years ago

7 0

The answer to your question is skew lines

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The model represents the equation x - 1 < -4. What is the solution for the inequality?

Step2247 [10]

Answer:

any number equal to or less than -4

Step-by-step explanation:

x - 1 < -4

< represents that -4 is a larger number than x - 1. we know that -4 is a negative number, so anything below -4 would be less. x = -4 would be a good solution, but the reality is that any number equal to or less than -4 is the correct answer. I hope this helps!

7 0

2 years ago

Which table represents an exponential function?

White raven [17]

Answer:

2nd one is the answer if it's the question

3 0

2 years ago

Help me plz I'm doing a test and I need help lol

My name is Ann [436]

Answer:

BC = BA

Step-by-step explanation:

6 0

2 years ago

A truck makes a trip of 14 hours, what was the speed if it traveled 2438 km?

lesantik [10]

Answer:

174

Step-by-step explanation:

I divided 2438 by 14 and got 174 mph.

6 0

3 years ago

Read 2 more answers

The marketing director of a large department store wants to estimate the average number of customers who enter the store every f

Dahasolnce [82]

Answer:

$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and $t_{\alpha /2}$ by 2.2621 we get:

$49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326$

Finally, the interval values that she get is:

$36.5674\leq x'\leq61.4326$

8 0

2 years ago