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2 years ago

MO bisects ∠LMN, m∠NMO =6x-20,and 2x+36. Solve for x and find m∠LMO

Mathematics

1 answer:

guapka [62]2 years ago

7 0

Answer:

fwe

Step-by-step explanation:

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Assume that the empirical rule is a good model for the time it takes for all passengers to board an airplane. The mean time is 4

slamgirl [31]

Answer:

The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 48, \sigma = 4$ .

Which interval describes how long it takes for passengers to board the middle 95% of the time?

This is between the 2.5th percentile and the 97.5th percentile.

So this interval is the value of X when Z has a a pvalue of 0.025 and the value of X when Z has a pvalue of 0.975

Lower Limit

Z has a pvalue of 0.025 when $Z = -1.96$ . So

$Z = \frac{X - \mu}{\sigma}$

$-1.96 = \frac{X - 48}{4}$

$X - 48 = -1.96*4$

$X = 40.16$

Upper Limit

Z has a pvalue of 0.975 when $Z = 1.96$ . So

$Z = \frac{X - \mu}{\sigma}$

$1.96 = \frac{X - 48}{4}$

$X - 48 = 1.96*4$

$X = 55.84$

The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.

7 0

3 years ago

Under her cell phone plan, Peyton pays a flat cost of $48 per month and $3 per gigabyte. She wants to keep her bill at $60.60 pe

Ad libitum [116K]

Answer: 4.20 gigabyte

Step-by-step explanation:

Based on the information given in the question, the equation that can be used to determine the the number of gigabytes of data Peyton can use while staying within her budget will be:

48 + 3g = 60.60

3g = 60.60 - 48

3g = 12.60

g = 12.60/3

= 4.20

The number of gigabytes of data is 4.20g

4 0

2 years ago

Read 2 more answers

How many decimal places do you move to change meters to centimeters?

vlada-n [284]

1 meter = 100 cm, so 10 m = 1000 cm and
0.1 m = 0.01
To change meter into centimeter you MULTIPLY the meters by 100
And to convert cm into meter you DIVIDE the cm by 100

8 0

3 years ago

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

3 years ago

The box and whisker plot below shows the numbers of text messages received in one day by students in the seventh and eighth grad

qaws [65]

They overlap at 22-26

6 0

3 years ago

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