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3 years ago

Answer and explanation please, ill give brainliest.

Mathematics

2 answers:

Minchanka [31]3 years ago

5 0

Answer: D

Explanation:

It’s a plot it’s asking how much of one thing

ratelena [41]3 years ago

3 0

Answer:

D, 4/8

Step-by-step explanation:

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The pH of a solution is measured eight times by one operator using the same instrument. She obtains the following data: 7.15, 7.

Nikolay [14]

Answer:

(a)7.185

(b)0.000371

(d)Sample variance and Sample standard deviation

Step-by-step explanation:

We are given the following sample data.

7.15, 7.20, 7.18, 7.19, 7.21, 7.20, 7.17, 7.18.

(a) Calculate the sample mean. Round your answer to 3 decimal places.

The formula for sample mean =

Sum of terms/Number of terms

= 7.15 + 7.20+ 7.18+7.19+7.21+7.20+7.17+ 7.18/8

= 57.48/8

= 7.185

(b) Calculate the sample variance. Round your answer to 6 decimal places.

The formula for sample Variance =

(x - mean)²/n - 1

n = 8

Sample variance = (7.15 - 7.185)² + (7.20-

7.185)²+ (7.18 -7.185)² + (7.19 -7.185)² + (7.21 - 7.185)² + (7.20 - 7.185)² + (7.17 - 7.185)² + (7.18 - 7.185)² /8 - 1

= 0.001225 + 0.000225+ 0.000025+ 0.000025 + 0.000625 + 0.000225 + 0.000225 + 0.000025/7

= 0.0026/7

= 0.000371428571

Approximately to 6 decimal places = 0.000371

The formula for Sample Standard Deviation = √Sample variance

= √0.000371428571

= 0.01927248223

To 3 decimal places = 0.019

(d) Which of the following are potential major sources of variability in the experiment

Sample variance

Sample standard deviation

4 0

3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

Help me with this easy problem If you answer correctly you will get brainliest and I wll draw you something. whaterver you like!

hjlf

Answer: 620

Step-by-step explanation:

$40 x 5.5 hrs = $220 This means that she earned $220 tutoring the first student.

$40 x 10 hrs = $400 She earned $400 tutoring the second student.

$220 + $400 = $620 She earned $620 in total that week.

7 0

2 years ago

Read 2 more answers