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3 years ago

Name the minor arc and find it's measure.

Mathematics

2 answers:

Vladimir79 [104]3 years ago

6 0

Based on the given figure, the minor arc is AB and its measure is 115°.

Minor arc is the arc that is less than half of the circle or less than 180°.
Major arc is the arc that is more than 180°. In this case, the major arc is ADB. Its measure is the difference of 360° and the minor arc, 245°

choli [55]3 years ago

5 0

Answer:

1.c. 72cm^2

2.c. m<1=36,m<2=72

3.b.7/2 and 49/4

4.c. 247.7 in^2

5.b. AB;115

6.a. 30 in

7.b. 10.89m^2

8.d. 11.7 m^2

9.(270pie+81/3)m^2

10.d. 20.73

Step-by-step explanation:

You might be interested in

Search cos jk and tan jk knowing that sin jk = 8/3

aleksley [76]

Answer:

cosjk = √55 i/3

tanjk = 8/√55 i

Step-by-step explanation:

Given

sin jk = 8/3

According to SOH CAH TOA

Sin theta = opposite/hypotenuse = 8/3

Opposite = 8

hypotenuse = 3

Get the adjacent using the pythagoras theorem

hyp² = opp²+adj²

adj² = hyp² - opp²

adj² = 3² - 8²

adj² = 9-64

adj² = -55

adj = √-55

adj = √55 i (i = √-1)

Get cosjk

cosjk = adj/hyp

cosjk = √55 i/3

Get tanjk

tanjk = opp/adj

tanjk = 8/√55 i

3 0

3 years ago

Perform the indicated operation.

Verizon [17]

Answer:

The answer is B. 298? im not sure if it was supposed to be ≈ or not

6 0

3 years ago

2-2x2-(39 divided by 1)+2+9

artcher [175]

-30 is the answer for your question

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.

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

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