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3 years ago

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In a statistics class, 10 scores were randomly selected with the following results: 74, 73, 77, 77, 71, 68, 65, 77, 67, 66. What

is the 65th percentile?

2 answers:

enyata [817]3 years ago

6 0

Answer:The 65th percentile is 74

Step-by-step explanation:

The total number of scores sampled is 10. So n = 10

We want to determine the 65th percentile. First step is to rearrange the scores in an increasing order. It becomes

65, 66, 67, 68, 71, 73, 74, 77, 77, 77

Next step is to determine the index by multiplying 65% by n. It becomes

65/100×10 = 6.5. Rounding up to the nearest whole number, it becomes 7

The 65th percentile will be the 7th term from the left. It becomes

74

julia-pushkina [17]3 years ago

4 0

Answer:

The 65th percentile is between 73 and 74

Step-by-step explanation:

Rearrange the ungrouped data in ascending order: 65,66,67,68,71,73,74,77,77,77

The 65th percentile is 65/100 of N ( N = 10) = 65/100 × 10 = 6.5

The 65th percentile is the 6.5th term. The 6th term is 73 and the 7th term is 74, the 6.5th term is between the 6th and 7th term

The 65th percentile is between 73 and 74

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iren2701 [21]

x2 - 12x - 12 = 0

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Hence, the answer is (B).

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3 years ago

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Step-by-step explanation:

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2 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Can someone help me with is one

ch4aika [34]

Answer:

C is your answer

Step-by-step explanation:

7 0

3 years ago

A baseball stadium sells game tickets for $8 each and has a capacity of 200 people. The function M(x) = 8x represents the amount

ollegr [7]

The domain of the function is whole numbers less than or equal to<em> </em>200

The given parameters;

cost of the game tickets = $8 per spectator

maximum capacity of the stadium = 200

The function for the amount of money the baseball stadium makes from ticket sales is given as;

M(x) = 8x

where;

x is the number of spectators

Since the maximum capacity of the stadium is 200, the number of spectators should be less or equal to 200.

x ≤ 200

Thus, the domain of the function is whole numbers <em>less than</em> or <em>equal to </em>200

Learn more here: brainly.com/question/16892163

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2 years ago