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2 years ago

Determine if JK and LM are parallel ,perpendicular, or neither. J(13,-5) K(2,6) L(-1,-5) M(-4,-2)

Mathematics

1 answer:

barxatty [35]2 years ago

7 0

Answer:

The lines are parallel

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Remember that

If two lines are parallel, then their slopes are the same

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

step 1

Find the slope JK

we have

$J(13,-5),K(2,6)$

substitute the values

$m=\frac{6+5}{2-13}$

$m=\frac{11}{-11}$

$m_J_K=-1$

step 2

Find the slope LM

we have

$L(-1,-5),M(-4,-2)$

substitute the values

$m=\frac{-2+5}{-4+1}$

$m=\frac{3}{-3}$

$m_L_M=-1$

step 3

Compare the slopes

$m_J_K=-1$

$m_L_M=-1$

$m_J_K=m_L_M$

therefore

The lines are parallel

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What proportion of these candy bars have more than 300 calories per serving? (Round your answer to two decimal places.)

Hatshy [7]

Answer: 0.12

Step-by-step explanation:

There are 65 candy bars. Out of that;

2 candy bars have 300 to 350 calories

1 candy bar has 350 to 400 calories

4 candy bars have 400 to 450 calories

1 candy bar has 450 to 500 calories

The total proportion out of 65 candy bars that have over 300 calories is;

= ( 2 + 1 + 4 + 1) / 65

= 8/65

= 0.12

3 0

2 years ago

The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac

Alina [70]

Answer:

a) $P(6.4$

b) $a=7 +1.036*0.333=7.345$

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =7,\sigma =1)$

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

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

$\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})$

Solution to the problem

Part a

(a) the probability that the mean life of a random sample of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333$

We want this probability:

$P(6.4$

Part b

b) The value of x to the right of which 15% of the means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

$P(\bar X>a)=0.15$ (a)

$P(\bar X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.036$

And if we solve for a we got

$a=7 +1.036*0.333=7.345$

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

8 0

2 years ago

What is the common multiple of 2,960 6,400 and 2000 with full process

Mamont248 [21]

Answer:

1184000

Step-by-step explanation:

lcm(2960,6400,2000)

prime factor:

$2960=(2^4)(5)(37)\\6400=(2^8)(5^2)\\2000=(2^4)(5^3)$

Take the largest power of each factor:

$(2^8)(5^3)(37^1)\\=1184000$

3 0

1 year ago

The number of calls coming per minute into a hotel reservation center is Poisson random variable with mean 3.

natima [27]

Answer:

a) 0.05

b) 0.9826

c) 0.000039308

Step-by-step explanation:

a) $P_X(0) = \frac{e^{-3}3^0}{0!} = e^{-3} = 0.05$

b) For two minutes, the mean is doubled, hence it is 6. In order to calculate the probability of al least two calls arriving, we calculate first the probability of the complementary event: At most 1 call will arrive. For that probability, we need to sum the probabilities of 0 and 1.

$P_X(0) = e^{-6}$