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lara31 [8.8K]
3 years ago
15

Please help ill give brainliest, and explain the answer too please

Mathematics
1 answer:
Basile [38]3 years ago
3 0

Answer:

x^2+2

Step-by-step explanation:

since (x^2+2x+1) is negative from the subtraction symbol in front, the entire thing turns negative. The equation turns into 2x^2+2x+3-x^2-2x-1=...

if you solve this equation you will get x^2+2 by adding the like terms together

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6 y 3

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Based on the information from Harper's Index, 37 out of 100 adult Americans who did not attend college believe in extraterrestri
vazorg [7]

Answer: Yes, this indicate the  proportion of adult americans that attended college and believe in extraterrestrials is higher than the proportion that did not attend college.

Step-by-step explanation:

Since we have given that

Hypothesis:

H_0:p_1=p_2\\\\H_a:p_1>p_2

n₁ = 100

x₁ = 37

So, p_1=\dfrac{x_1}{n_1}=\dfrac{37}{100}=0.37

n₂ = 100

x₂ =47

So, p_2=\dfrac{47}{100}=0.47

At 0.01 level of significance, z = 2.33

So, test statistic value would be

z=\dfrac{p_1-p_2}{\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}}\\\\z=\dfrac{0.37-0.47}{\sqrt{\dfrac{0.37\times 0.63}{100}+\dfrac{0.47\times 0.53}{100}}}\\\\z=-1.44

Since, 2.33>-1.44

Hence, we will reject the null hypothesis.

Therefore, Yes, this indicate the  proportion of adult americans that attended college and believe in extraterrestrials is higher than the proportion that did not attend college

4 0
3 years ago
Giving brainliest!!!!
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An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen
Slav-nsk [51]

Answer:

Median = 90.4

Q_1 = 88.6

Q_3 = 92.2

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

Smallest = 83.4

Highest = 100.3

Next, group each dataset base on common whole numbers.

So, we have:

83.4

84.3\ 84.3

85.3

86.7\ 86.7\ 86.7

87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9

88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9

89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9

90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9

91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\  91.6\ 91.6\ 91.8\ 91.8

92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7

93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7

94.2\ 94.2\ 94.4\ 94.7

96.1\ 96.5

98.8

100.3

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}

  \begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}

From the above plot,

n = 83

The median is calculated as:

Median = \frac{n+1}{2}th

Median = \frac{83+1}{2}th

Median = \frac{84}{2}th

Median = 42nd

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf  .4

So the median is:

Median = 90.4

The lower quartile (Q) is calculated as:

Q_1 = \frac{n+1}{4}th

Q_1 = \frac{83+1}{4}th

Q_1 = \frac{84}{4}th

Q_1 = 21st

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

Q_1 = 88.6

The upper quartile (Q3) is calculated as:

Q_3 = 3 * \frac{n+1}{4}th

Q_3 = 3 * \frac{83+1}{4}th

Q_3 = 3 * \frac{84}{4}th

Q_3 = 3 * 21th

Q_3 = 63rd

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

Q_3 = 92.2

8 0
3 years ago
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