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Lera25 [3.4K]
3 years ago
8

Solve the following math problems, round appropriately according to sig. fig. rules. 84.9 • 64

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

84.9*64=5433.6

3900/2.6=1500

1080/15.9=67.9245283019

2.86*8.77*200=5016.44

1500/64=23.4375

6007/5.5=1092.1818....

Step-by-step explanation:

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aleksandr82 [10.1K]

Answer:

A. 7- i

B. -20 -10i

C. 50

D. -1 +2i

Step-by-step explanation:

7 0
3 years ago
In the diagram, the measure of angle 8 is 124°, and the measure of angle 2 is 84°.
Darina [25.2K]

Answer:

(A) 56 on ed genuity

Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the
mafiozo [28]

Answer:

Porcentage of students score below 62 is close to 0,08%

Step-by-step explanation:

The rule

68-95-99.7

establishes:

The intervals:

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ] contains 68.3 % of all the values of the population

[ μ₀ - σ ,  μ₀ + σ]   contains 95.4 % of all the values of the population

[ μ₀ - 1,5σ ,  μ₀ + 1,5σ] contains 99.7 % of all the values of the population

In our case such intervals become

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ]   ⇒  [ 78 - (0,5)*8 , 78 + (0,5)*8 ]  ⇒[ 74 , 82]

[ μ₀ - σ ,  μ₀ + σ]  ⇒ [ 78 - 8 , 78 +8 ]   ⇒  [ 70 , 86 ]

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]  ⇒ [ 78 - 12 , 78 + 12 ]  ⇒ [ 66 , 90 ]

Therefore the last interval

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]    ⇒  [ 66 , 90 ]

has as lower limit 66 and contains 99.7 % of population, according to that the porcentage of students score below 62 is very small, minor than 0,15 %

100 - 99,7  = 0,3 %

Only 0,3 % of population is out of   μ₀ ± 1,5 σ, and by symmetry 0,3 /2 = 0,15 % is below the lower limit, 62 is even far from 66 so we can estimate, that the porcentage of students score below 62 is under 0,08 %

6 0
3 years ago
A Converse Statement is when you turn the statement into a double negative?
Misha Larkins [42]

Answer:

False

Step-by-step explanation:

This is because there is no such thing! A double negative becomes a positive! Plz give brainliest!

3 0
3 years ago
Read 2 more answers
Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx.
julsineya [31]
<span>since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
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</span>
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</span><span>pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]
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<span>cos(2t) = 2cos^2 - 1
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<span>1/sqrt(2) - (-1/sqrt(2) +1)
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3 0
3 years ago
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