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2 years ago

8

Isosceles triangle LMN is graphed with vertices L(0, 1), M(3, 5), and N(6,1). What is the slope of side LN? Negative StartFracti

on 4 over 3 EndFraction. 0 StartFraction 4 Over 3 EndFraction. 3

2 answers:

Mekhanik [1.2K]2 years ago

6 0

Answer:

the answer is 0

Step-by-step explanation:

Goshia [24]2 years ago

3 0

Answer:

B.0

Step-by-step explanation:

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If AB = 13, BC = 9, and CA = 17, list the angles of ABC in order from smallest to largest.

svp [43]

Across the largest side is going to be largest angle, and across the smallest side -smallest angle.

Sides:

BC = 9 < AB = 13 < CA = 17.

Angles that are across sides: A < C < B.

Answer is d- A C B.

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2 years ago

Find the z-scores that bound the middle 74% of the area under the standard normal curve.

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P (left) = 0.13

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At this P values, the z scores are approximately:

z score (left) = -2.22

<span>z score (right) = 1.13</span>

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1 year ago

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1 year ago

Read 2 more answers

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr

Gre4nikov [31]

Answer:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this: $P(-1.5 \leq Z \leq 1.5) = P(z$ Step-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(50,12)$

Where $\mu=50$ and $\sigma=12$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required like this:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this:

$P(-1.5 \leq Z \leq 1.5) = P(z$

4 0

1 year ago

One cube has an edge length 5 cm shorter than the edge length of the second cube. The volume of the smaller cube is 216 cm?. Wha

VLD [36.1K]

Answer:

The volume of the larger cube is $V=1,331\ cm^3$

Step-by-step explanation:

Let

x ---> the length of the smaller cube

y ---> the length of the larger cube

we know that

the volume of a cube is equal to

$V=b^3$

where

b is the length side of the cube

we have

$x=y-5$ ----> equation A

The volume of the smaller cube is 216 cm^3

so

substitute in the formula of volume

$216=x^3$

solve for x

$x=\sqrt[3]{216}\ cm$

$x=6\ cm$

<em>Find the value of y</em>

$x=y-5$

$6=y-5$

$y=6+5=11\ cm$

<em>Find the volume of the larger cube</em>

$V=y^3$

substitute the value of y

$V=11^3$

$V=1,331\ cm^3$

4 0

2 years ago