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3 years ago

5/7 × 4/3 ÷ 15/21 =

Mathematics

2 answers:

Montano1993 [528]3 years ago

6 0

Answer:

4/3

Step-by-step explanation:

kirill115 [55]3 years ago

4 0

Answer:

Your answer is 4/3

Step-by-step explanation:

or 1.3

or 1 1/3 hope this helps

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2 years ago

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Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering

Alla [95]

Answer:

a) The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

Step-by-step explanation:

a) To approximate this distribution we have to calculate the mean and the standard distribution.

The mean is the proportion p=0.85.

The standard deviation can be calculates as:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{100} }=0.04$

To calculate the probability that Jodi scores 78% or less on a 100-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.04} =-1.75$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) In this case, the number of questions is 250, so the standard deviation needs to be calculated again:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{250} }=0.02$

To calculate the probability that Jodi scores 78% or less on a 250-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.02} =-3.5$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

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2 years ago

Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea

lord [1]

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

$\int\limits^y_0 {kx^{2} } \, =1$

where y=23

Clearing k=0.00025

a) $Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5$

b) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2} =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6$

c) The function is equal to:

f(x)=k(1+2x)

$\int\limits^y_0 {k(1+2x)} \, =1$

where y=20

k=0.0024

$Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3$

d) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2} =198.4-176.89=21.51$

8 0

3 years ago

5/7 × 4/3 ÷ 15/21 =​

5/7 × 4/3 ÷ 15/21 =