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crimeas [40]
3 years ago
10

Anybody can help me with this ?

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
6 0

Answer:

Step-by-step explanation:

y=3x+1

y=3(3)+1

y=9+1

y=10

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Nady [450]

Answer:  b) -0.625

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

The formula for the z-score is:

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6 0
3 years ago
Give the following compound's base name.
kenny6666 [7]
Amswer the first one
3 0
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ArbitrLikvidat [17]

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Step-by-step explanation:

8 0
3 years ago
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1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
In Game 1, Emerson struck out 30 times in 90 times at bat. In Game 2, he struck out 40 times in 120 times at bat. In Game 3, Eme
VikaD [51]

Answer:

<u>Total number of strikeouts:</u>

  • 30 + 40 + 42 = 112

<u>Total number of at bats:</u>

  • 90 + 120 + 140 = 350

<u>Hits = times - strikeouts:</u>

  • 350 - 112 = 238

<u>Percentage:</u>

  • 238/352*100% = 67.61%

7 0
3 years ago
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