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dangina [55]
3 years ago
7

The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. What percent of the scores are

less than 81?
A. 84%

B. 13.5%

C. 50%

D. 16%

Mathematics
2 answers:
Lynna [10]3 years ago
4 0
A normal distribution with a mean of 74 and a standard deviation of 7.
Mean + 1 SD = 74 + 7 = 81
Less than 81 :  50 % + 34 % = 84 %
Answer:
A ) 84 % 
IceJOKER [234]3 years ago
4 0

Answer:

<em>84%</em><em> of the scores are less than 81.</em>

Step-by-step explanation:

Here,

The given distribution is in normal distribution.

Mean = 74

Standard deviation = 7

Approximately 68% of the data falls within one standard deviation of the mean (or between the mean -one times the standard deviation, and the mean + 1 times the standard deviation).

As 81 = 74+7 = μ+1σ

So, \dfrac{64}{2}=34\% of scores will be between 74 and 81.

As 74 is the mean, so 50% of score will lie below 74.

So total of 50+34=84% of the score will be less than 81.

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The answer is B) x^3-1 because it's the only one with the 3rd degree
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The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0°C at the freezing
PSYCHO15rus [73]

Answer:

-1.53 , 1.53

Step-by-step explanation:

It is basically asking you to find the percentiles. so for the bottom 6.3% you would type into your calculator 2nd, VARS, invNorm, area: .063, mean: 0, and Standard Deviation: 1 which gives you -1.53.

for the upper 6.3% you have to take 1-.063=.937 to get the upper percentile. now type into your calculator 2nd, VARS, invNorm, area: .937, mean: 0, and Standard Deviation: 1 which gives you 1.53.


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6 0
1 year ago
If
baherus [9]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: cos 330 = \frac{\sqrt3}{2}

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}

Proof LHS → RHS:

LHS                          cos 165

Double-Angle:        cos (2 · 165) = 2 cos² 165 - 1

                             ⇒ cos 330 = 2 cos² 165 - 1

                             ⇒ 2 cos² 165  = cos 330 + 1

Given:                        2 \cos^2 165  = \dfrac{\sqrt3}{2} + 1

                              \rightarrow 2 \cos^2 165  = \dfrac{\sqrt3}{2} + \dfrac{2}{2}

Divide by 2:               \cos^2 165  = \dfrac{\sqrt3+2}{4}

                             \rightarrow \cos^2 165  = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}

                             \rightarrow \cos^2 165  = \dfrac{2\sqrt3+4}{8}

Square root:             \sqrt{\cos^2 165}  = \sqrt{\dfrac{4+2\sqrt3}{8}}

Scratchwork:            \cos^2 165  = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2

                             \rightarrow \cos 165  = \pm \dfrac{\sqrt3+1}{2\sqrt2}

             Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

                             \rightarrow \cos 165  = - \dfrac{\sqrt3+1}{2\sqrt2}

LHS = RHS \checkmark

4 0
3 years ago
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olga55 [171]

Answer:

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Step-by-step explanation:

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7 0
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tankabanditka [31]
ANSWER-

2x(2x^2-4-x^5)

EXPLANATION-

GCF is 2 then factor
8 0
3 years ago
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