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lutik1710 [3]
4 years ago
5

Which sentence represents a quadratic equation that can be expressed in standard form?

Mathematics
1 answer:
wariber [46]4 years ago
7 0

Answer:

It's c.

Step-by-step explanation:

C) is (x - 1)^2 = 36

x^2 - 2x + 1 = 36

x^2 - 2x - 35 = 0

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2x-3y = 2y - x then find the value of ×:y
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If 2x+y=2y−x, then the value of y in terms of x can be

3x

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The amount spent (in billions of dollars) for ads online is shown. (The numbers for 2016 through 2019 are projected numbers.) Dr
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Find x. Round your answer to the nearest tenth of a degree.
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Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple
polet [3.4K]

Answer:

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Step-by-step explanation:

We are given that a simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car.

Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                             P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  }  ~ N(0,1)

where, \hat p_1 = sample proportion of small cars that were totaled = \frac{8}{12} = 0.67

\hat p_2 = sample proportion of large cars that were totaled = \frac{5}{15} = 0.33

n_1 = sample of small cars = 12

n_2 = sample of large cars = 15

p_1 = population proportion of small cars that are totaled

p_2 = population proportion of large cars that were totaled

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

So, 95% confidence interval for the difference between population population, (p_1-p_2) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                    of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  } < p_1-p_2 < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  } ) = 0.95

<u>95% confidence interval for</u> p_1-p_2 = [(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  } , (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }  }]

= [(0.67-0.33)-1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} }  } , (0.67-0.33)+1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} }  }]

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].

6 0
4 years ago
1. A cone is 8cm high and has a base diameter of 12cm.its slant height is a.6cm b.8cm c.10cm d.12cm
Anit [1.1K]

Answer:

10

Step-by-step explanation:

it is Pythagoras theorem

6*6=36

8*8=64

64+36=100

square root of 100 is 10

4 0
3 years ago
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