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3 years ago

Please help! I will give out brainlist.

Mathematics

1 answer:

Ainat [17]3 years ago

7 0

Answer:

Step-by-step explanation:

This is a Pythagorean Theorem Question

c^2 = a^2 + b^2

c= sqrt(5)

a = x

b = x This is an isosceles right triangle. The vertical side and the horizontal side are equal.

x^2 + x^2 = sqrt(5)^2

2x^2 = 5 Divide both sides by 2

x^2 = 5/2 Take the sqrt of both sides.

√x^2 = √5 / √2 Multiply right side by √2 on both top and bottom

x = √5*√2 / √2*√2

x = √10 /2

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POSSIBLE POINTS: 10

Nadusha1986 [10]

Answer:

5 + 3t

Step-by-step explanation:

t = amount Tanya has

Lucy = $5 more than 3x of what Tanya has

5 0

3 years ago

An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e

uranmaximum [27]

Answer:

The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938

Step-by-step explanation:

Given:

Here Section 1 students = 20

Section 2 students = 30

Here there are 15 graded exam papers.

(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070

(b) Here if x is the number of students copies of section 2 out of 15 exam papers.

here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15

Then,

Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798

(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)

so,

Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938

Note : Here the given distribution is Hyper-geometric distribution

where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.

7 0

3 years ago

Mary took 8 tests in science and received the following scores: 87,60,76,92,63,91,88,75

MariettaO [177]

Answers:

mean = 79
mode = none
median = 81.5

=================================================

Explanation:

To get the mean, you add up the numbers and then divide by n = 8, since there are 8 scores

Adding the scores gets us: 87+60+76+92+63+91+88+75 = 632

Divide that over n = 8 to get 632/n = 632/8 = 79

The mean is 79

You have the correct answer. Nice work.

---------------

The mode is the most frequent value.

In this data set, we don't have any repeated values. Each unique number is listed one time only. So that tells us we don't have a mode here.

---------------

To get the median, we need to sort the items from smallest to largest

{87,60,76,92,63,91,88,75} sorts to {60,63,75,76,87,88,91,92}

Because we have n = 8 values, which is an even number, this tells us that the median is between slot n/2 = 8/2 = 4 and slot 5

The values 76,87 are in slots four and five in that order. Add them up and divide by 2: (76+87)/2 = 163/2 = 81.5 is the median

3 0

3 years ago

Use the quadratic formula to find the solutions to the quadratic equation below 2x^2-5x+5=0

Serggg [28]

Answer:

x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4

Step-by-step explanation:

Solve for x:

2 x^2 - 5 x + 5 = 0

Hint: | Using the quadratic formula, solve for x.

x = (5 ± sqrt((-5)^2 - 4×2×5))/(2×2) = (5 ± sqrt(25 - 40))/4 = (5 ± sqrt(-15))/4:

x = (5 + sqrt(-15))/4 or x = (5 - sqrt(-15))/4

Hint: | Express sqrt(-15) in terms of i.

sqrt(-15) = sqrt(-1) sqrt(15) = i sqrt(15):

Answer: x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4

5 0

4 years ago

Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this

finlep [7]

Answer:

The answer is " $0.90>p>0.75.$ "

Step-by-step explanation:

$\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ -6 36 \ \ \ \ \ \ \ \ \ \ \ 0.092308\\\\$

$roof \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 408 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ 18 \ \ \ \ \ \ \ \ \ \ \ 324 \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \ 70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \ 49 \ \ \ \ \ \ \ \ \ \ \ 0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 \ \ \ \ \ \ \ \ \ \ \ 70 \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 0.057143\\\\$

$vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.257143\\\\$

$\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.028571\\\\\text{roof vestigial} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2\\\\$

$\text{cinnabar} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8 \\\\$

$Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.965934$

Eight phenotypes were present.

Df is provided also by a number of phenotypes -1 The degree of freedom

$\to df = 8-1= 7$

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

7 0

3 years ago