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3 years ago

Chuck drew an angle measuring 148°.

Mathematics

2 answers:

Ghella [55]3 years ago

5 0

Im going to guess B., obtuse. i know botuse angles go from 100-160 degrees. sorry if im wrong, but im sure im not.

Wewaii [24]3 years ago

3 0

B. obtuse angle

obtuse angle is angle that is wider than 90 and less than 180

thus 148 is an obtuse ^^

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The sum of the measures of two complementary angles exceeds the difference of their measures by 86. Find the measure of the larg

USPshnik [31]

You will need to know that the sum of "two complementary angles" is 90
therefore
Let x = one angle
then
90-x = second angle
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Now, we construct our equation based on the sentence:
"The sum of the measures of two complementary angles exceeds the difference of their measures by 86°."
x + 90-x = x-(90-x)+86
90 = x-90+x+86
90 = 2x-90+86
90 = 2x-4
94 = 2x
47 degrees = x (first angle)
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Second angle:
90-x = 90-47 = 43 degrees (second angle) hope it helps

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3 years ago

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The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db; which is to say, this may not b

WINSTONCH [101]

Answer:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

Step-by-step explanation:

The problem is incomplete. The questions are:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

$M-z\sigma/\sqrt{n}\leq\mu\leq M-z\sigma/\sqrt{n}\\\\47-2.58*10/\sqrt{75} \leq\mu\leq47+2.58*10/\sqrt{75}\\\\47-2.98\leq\mu\leq47+2.98\\\\44.02\leq\mu\leq 49.98$

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

$E=z*\sigma/\sqrt{n}=1.64*10/\sqrt{75}=1.89$

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.

The values of db for these critical values are:

$X_1=M+z_1*\sigma/\sqrt{n}=47+(-2.33)*10/\sqrt{81}=47-2.59=44.41\\\\\\ X_2=M+z_2*\sigma/\sqrt{n}=47+(2.33)*10/\sqrt{81}=47+2.59=49.59$

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