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3 years ago

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The scores for a certain test of intelligence are normally distributed with mean 85 and standard deviation 11. Find the 90th per

centile of these scores.

1 answer:

jonny [76]3 years ago

5 0

Answer:

99.1 is the 90th percentile of these scores.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 85

Standard Deviation, σ = 11

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.90

$P( X < x) = P( z < \displaystyle\frac{x - 85}{11})=0.90$

Calculation the value from standard normal z table, we have,

$P(z \leq 1.282) = 0.90$

Putting values, we get,

$\displaystyle\frac{x - 85}{11} = 1.282\\\\x = 99.1$

Thus, 99.1 is the 90th percentile of these scores.

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The ordered pairs that satisfy this function are (-3,-1) and (3,3)

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Find the equation of the line that passes through the points (-2, 3) and (1, -6). Write your answer in Standard Form.

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Slope = (-6 - 3)/(1+2) = -9/3=-3
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3 years ago

You have 100 cm of string which can be cut in one place (or not cut at all) and then formed into a circle and a square (or just

Ne4ueva [31]

Answer:

44cm for minimum area and 0 for maximum area (circle)

Step-by-step explanation:

Let's C be the circumference of the circle and S be the circumference of the square. If we cut the string into 2 pieces the total circumferences would be the string length 100cm.

S + C = 100 or S = 100 - C

The side of square is S/4 and radius of the circle is $\frac{C}{2\pi}$

So the area of the square is

$A_S = \frac{S^2}{4^2} = \frac{S^2}{16}$

$A_C = \pi\frac{C^2}{(2\pi)^2} = \frac{C^2}{4\pi}$

Therefore the total area is

$A = A_S + A_C = \frac{S^2}{16} + \frac{C^2}{4\pi}$

We can substitute 100 - C for S

$A = \frac{(100 - C)^2}{16} + \frac{C^2}{4\pi}$

$A = \frac{100^2 - 200C + C^2}{16} + \frac{C^2}{4\pi}$

$A = 625 -12.5C + \frac{C^2}{16} + \frac{C^2}{4\pi}$

$A = 625 -12.5C + C^2(\frac{1}{16} + \frac{1}{4\pi})$

To find the maximum and minimum of this, we can take the first derivative and set that to 0

$A^{'} = -12.5 + 2C(\frac{1}{16} + \frac{1}{4\pi}) = 0$

$C(\frac{1}{8} + \frac{1}{2\pi}) = 12.5$

$C \approx 44 cm$

If we take the 2nd derivative:

$A^{''} = \frac{1}{8} + \frac{1}{2\pi} > 0$

We can see that this is positive, so our cut at 44 cm would yield the minimum area.

The maximum area would be where you not cut anything and use the total string length to use for either square or circle

if C = 100 then $A_C = \frac{C^2}{4\pi} = \frac{100^2}{4\pi} = 795.77 cm^2$

if S = 100 then $A_S = \frac{S^2}{16} = \frac{100^2}{16} = 625 cm^2$

So to yield maximum area, you should not cut at all and use the whole string to form a circle

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3 years ago

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luda_lava [24]

Answer:

$\angle m=90^o$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The measurement of the external angle is the half-difference of the arches that comprise

so

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Remember that the sum of the major arc and a minor arc is equal to 360 degrees

we have

$major\ arc\ LN=270^o$ ----> is given

so

$minor\ arc\ LN=360^o-270^o=90^o$

substitute the values

$\angle m=\frac{1}{2}(270^o-90^o)=90^o$

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Step-by-step explanation:

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