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JulsSmile [24]
2 years ago
14

Complete the given ordered pairs for the equation. y = 2x + 1 (5, ) (0, ) (-2, )

Mathematics
1 answer:
VikaD [51]2 years ago
6 0

Answer:

(5,11)

(0,1)

(-2,-3)

Step-by-step explanation:

y=2(5)+1

y=10+1

y=11

so... (5,11)

y=2(0)+1

y=0+1

y=1

so... (0,1)

y=2(-2)+1

y=-4+1

y=-3

so... (-2,-3)

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19/5 as a mixed number
Feliz [49]
The answer would be 3 and 4 over 5
8 0
3 years ago
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Sean tossed a coin off a bridge into the stream below. The path of the coin can be represented by the equation 2 h tt = − 16t^2+
tekilochka [14]

Answer:

It will take 5.61 seconds for the coin to reach the stream.

Step-by-step explanation:

The height of the coin, after t seconds, is given by the following equation:

h(t) = -16t^{2} + 72t + 100

How long will it take the coin to reach the stream?

The stream is the ground level.

So the coin reaches the stream when h(t) = 0.

h(t) = -16t^{2} + 72t + 100

-16t^{2} + 72t + 100 = 0

Multiplying by (-1)

16t^{2} - 72t - 100 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

16t^{2} - 72t - 100 = 0

So

a = 16, b = -72, c = -100

\bigtriangleup = (-72)^{2} - 4*16*(-100) = 11584

t_{1} = \frac{-(-72) + \sqrt{11584}}{2*16} = 5.61

t_{2} = \frac{-(-72) - \sqrt{11584}}{2*16} = -1.11

Time is a positive measure, so we take the positive value.

It will take 5.61 seconds for the coin to reach the stream.

3 0
3 years ago
One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probab
Andrews [41]
You only need to consider the situations where 10 or 11 of the babies are girls, then subtract those probabilities from 1.  This will give probability that any other number up to 9 of the babies are girls.

Use binomial theorem.
P(x=k) = (nCk) p^k (1-p)^{n-k}

n = 11
k = 10,11
p = 1/2

P(x=10) = 11 (\frac{1}{2})^{11} = \frac{11}{2048} \\  \\ P(x=11) = 1(\frac{1}{2})^{11} = \frac{1}{2048} \\  \\ P(x \leq 9) = 1 - \frac{11}{2048} - \frac{1}{2048} \\  \\ P(x \leq 9)=\frac{2036}{2048} = \frac{509}{512}
4 0
3 years ago
Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico
IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

                                  P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean nicotine content = 27.3 milligrams

            s = sample standard deviation = 2.8 milligrams

            n = sample of cigarettes = 9

            \mu = true mean nicotine content

<em>Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>Part I</u> : So, 99% confidence interval for the population mean, \mu is ;

P(-3.355 < t_8 < 3.355) = 0.99  {As the critical value of t at 8 degree

                                      of freedom are -3.355 & 3.355 with P = 0.5%}  

P(-3.355 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 3.355) = 0.99

P( -3.355 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 3.355 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X-3.355 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+3.355 \times {\frac{s}{\sqrt{n} } } ) = 0.99

<u />

<u>99% confidence interval for</u> \mu = [ \bar X-3.355 \times {\frac{s}{\sqrt{n} } } , \bar X+3.355 \times {\frac{s}{\sqrt{n} } } ]

                                          = [ 27.3-3.355 \times {\frac{2.8}{\sqrt{9} } } , 27.3+3.355 \times {\frac{2.8}{\sqrt{9} } } ]

                                          = [27.3 \pm 3.131]

                                          = [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

<u>Part II</u> : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

                                  P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean nicotine content = 24.9 milligrams

            s = sample standard deviation = 2.6 milligrams

            n = sample of cigarettes = 9

            \mu = true mean nicotine content

<em>Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

So, 98% confidence interval for the population mean, \mu is ;

P(-2.896 < t_8 < 2.896) = 0.98  {As the critical value of t at 8 degree

                                       of freedom are -2.896 & 2.896 with P = 1%}  

P(-2.896 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.896) = 0.98

P( -2.896 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.896 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.896 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.896 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u />

<u>98% confidence interval for</u> \mu = [ \bar X-2.896 \times {\frac{s}{\sqrt{n} } } , \bar X+2.896 \times {\frac{s}{\sqrt{n} } } ]

                                          = [ 24.9-2.896 \times {\frac{2.6}{\sqrt{9} } } , 24.9+2.896 \times {\frac{2.6}{\sqrt{9} } } ]

                                          = [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0
2 years ago
Which of the following is equivalent to 2:1​
Len [333]

Answer:

use m a t h w a y

Step-by-step explanation:

put it all together and they will give you opitions

7 0
3 years ago
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