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son4ous [18]
3 years ago
12

1.4 has 2 significant digits as a result of which rule? a. All nonzero numbers are significant. b. Zeros between nonzero digits

are significant. c. Final zeros after the decimal point are significant. d. Zeros used solely for spacing the decimal point are not significant.
Mathematics
1 answer:
marishachu [46]3 years ago
6 0

Answer:

All non zero digits count in significant digits.

(a) is correct option.

Step-by-step explanation:

Given that,

1.4 has 2 significant digits.

We know that,

Significant digits :

All non zero digits count in significant digits.

All zero which is present in between two significant digits it is also significant.

Leading zero is not count as significant.

Trailing zero is count as significant.

We need to find 1.4 has 2 significant digits as a result of which rule

Using given rules

1.4 has 2 significant digits it is follows the rule of all non zero digits count in significant digits.

Hence, All non zero digits count in significant digits.

(a) is correct option.

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If (x-10) is a factor of the polynomial P(x), which of the following do we know to be true?
Anon25 [30]

Answer:

Option 3, p(0) = -10

Option 4, p(10) = 0

Step-by-step explanation:

<u>Step 1:  Check</u>

x - 10 + 10 = 0 + 10

<em>x = 10</em>

f(0) = 0 - 10 = -10

x = -10

<em>p(10) = 0</em>

Answer:  Option 3, p(0) = -10, Option 4, p(10) = 0

4 0
3 years ago
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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
IRINA_888 [86]

Answer:

<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>

<em>P( 73 < x < 84 ) = 0.02145 = 2.1 %</em>

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

<u><em>Explanation</em></u>:-

Given mean of the Population 'μ ' = 51 months

Standard deviation of the Population 'σ' = 11 months

Let 'X' be the random variable of Normal distribution

<em>Let    'X'  = 73</em>

<em></em>Z = \frac{x-mean}{S.D} = \frac{73-51}{11} = 2<em></em>

<em>Let  'X' = 84</em>

<em></em>Z = \frac{84-51}{11} = 3<em></em>

<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>

<em>P( 73 < x < 84 )      = P( 2 < Z < 3)</em>

<em>                               = P( Z<3) - P( Z <2)</em>

<em>                              =  0.5 + A(3) - ( 0.5 + A(2))</em>

<em>                             = A(3) - A( 2)</em>

<em>                             = 0.49865 - 0.4772     ( From Normal table)</em>

<em>                             = 0.02145</em>

<em>  P( 73 < x < 84 ) = 0.02145</em>

<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>

<em>P( 73 < x < 84 ) = 0.02145 = 2.1 %</em>

7 0
3 years ago
Which of the following properties is shown in the question below?
trapecia [35]

Answer:

C. Associative property of multiplication

Step-by-step explanation:

The associative property of multiplication states that the way in which factors are grouped in a multiplication problem does not change the product.

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2 years ago
Which choices listed below indicate that a linear model is not the best fit for a dataset? Choose all that apply.
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<span>The following indicate that a linear model is not the best fit for a dataset:
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• Scatterplot shows a curve pattern. 
• Residual plot shows no pattern. 
• Correlation coefficient is close to 1 or –1. 

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6 0
3 years ago
Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights
Fed [463]

Answer:

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

  • Given that,  A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

  • Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

  • P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.

Next, multiply that by the sample size of 732.

  • Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171  

 

  • For part b, use the same method except x is now 1591.    

Z = (1581 - 3311)/860 = -2    

  • P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

 

  • For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2  

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

  approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.

5 0
2 years ago
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