Ask question

Ask question

All categories

3 years ago

7

When finding percentiles, if the locator L is not a whole number, one procedure is to interpolate so that a locator of 23.75,

for example, leads to a value that is three fourths of the way between the 23rd and 24th scores. Use this method of interpolation to find for the set of test scores below. 51 54 64 68 72 74 76 83 94 94 99

1 answer:

inessss [21]3 years ago

7 0

Answer:

The answer is "94"

Step-by-step explanation:

Given data:

51,54,64,68,72,74,76,83,94,94,99

Find:

$P_{75}=?P_{75}= (\frac{75(n+1)}{100})^{th} \ \ observation\\\\$

$= (\frac{75(11+1)}{100})^{th} \ \ observation\\\\= (\frac{75(12)}{100})^{th} \ \ observation\\\\= (\frac{75 \times 12}{100})^{th} \ \ observation\\\\= (\frac{900}{100})^{th} \ \ observation\\\\= 9^{th} \ \ observation\\\\= 9^{th} \ number \ is = 94$

You might be interested in

What is the answer to this problem 1/2=q 2/3?

Angelina_Jolie [31]

3/2 = 2q
q = 3/2 ÷ 2
q = 3/2 × 1/2
q = 3/4

6 0

3 years ago

Share £180 in the 4:5

marissa [1.9K]

4 + 5 = 9
£180 / 9 = £20

£20 x 4 = £80
£20 x 5 =£100

Answer:
£80 : £100

hope it helped :)

4 0

3 years ago

When y = 7, what is the value of x?

GREYUIT [131]

Answer:-1 as said in the bottom of the equation

Step-by-step explanation: it is at the bottom of the equation if wrong sorry.

3 0

3 years ago

If Dora was born in 1981 and her son was born 25 years after that what is an equation that could help solve that

Gnom [1K]

1981+25
Since she was born in the year 1981, and her son was born 25 years later so you add 25

4 0

3 years ago

Read 2 more answers

D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

3 years ago