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3 years ago

15/3 = 5 4_7 + 5 2_3 b=?

Mathematics

2 answers:

OLEGan [10]3 years ago

5 0

Answer:

válla si son brutos está bien fácil esa pregunta lo que pasa es que estoy haciendo mi trabajo

Sonbull [250]3 years ago

3 0

Answer:

0 < 5 < 54/7 < 52/3

Step-by-step explanation

0/21 < 105/21 < 162/21 < 364/21

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I need help finding the answers someone plz help

Harlamova29_29 [7]

Happy New Year from MrBillDoesMath!

Answer:

Proof by ASA congruence postulate. See below

Discussion:

Fact 1 : angle A = angle T (given)

Fact 2: The angles on both sides of point X are equal as vertical angles

are equal.

From these facts it follows that angle M = angle H (as all plane triangles have 180 degrees). Also AM = TH (given) so

In the left triangle In the right triangle

(angle M, side AM, angle A) = (angle H, side TH, angle T)

Hence the triangles have two congruent angles, and congruent sides included between the angles, so they are congruent by ASA.

Thank you,

MrB

8 0

3 years ago

How do i do this i dont know were to begin

loris [4]

Multiply everything by 4 so you get rid of the fraction

6 0

3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$