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3 years ago

P(A) = 0.44 What are the odds for A?

Mathematics

2 answers:

damaskus [11]3 years ago

8 0

$P(A)=0.44=\frac{44}{100}=\frac{11}{25}\\\\11\ odds\ on\ 25$

Kisachek [45]3 years ago

5 0

0,44 = 44 / 100

44 odds on 100

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25 is what percent less than 35

mars1129 [50]

Difference between 25 and 35 = 35 -25
   = 10
Then
percentage by which 25 is less than 35 =(10/35) * 100
   = (2/7) * 100
   = 200/7
   = 28.57 percent
So 25 is 28.57% less than 35.This is the only clear way of solving these kind of problems. I hope you have understood the method used to solve this problem. Hopefully you can do such type of problems without needing any help.

8 0

3 years ago

Read 2 more answers

Find the output, kkk, when the input, ttt, is -7−7minus, 7. k = 10t-19k=10t−19

Sati [7]

The output k at input t = -7 is -89

Step-by-step explanation:

In order to find the value of a expression on given number, we put the number in given expression in place of independent variable

So,

Given expression is:

$k = 10t-19$

It is given that the value of t has to be -7

so putting t=-7 in the expression

$k = 10(-7)-19\\k = -70-19\\k = -89$

Hence,

The output k at input t = -7 is -89

Keywords: Expressions, input, output

Learn more about expressions at:

#LearnwithBrainly

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3 years ago

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Evaluate: (5x) expression 2 - 3 if x=2 Help please

charle [14.2K]

Assuming that you mean 5x(2-3) for your equation, you would just have to plug 2 into where x is. This makes your equation 5*2(2-3). This gives you 10(-1) because 5*2=10, and 2-3=-1. Then simply multiply the two to get:

-10

Hope this helped!

4 0

3 years ago

A study of pollutants showed that certain industrial emissions should not exceed 2.3 parts per million. You believe a particular

Musya8 [376]

Answer:

$t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5$

The degrees of freedom are given by:

$df=n-1=9-1=8$

And the p value would be:

$p_v =P(t_{(8)}>7.5)=0.0000346$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

Step-by-step explanation:

Information provided

$\bar X=3.3$ represent the sample mean in ppm

$s=0.4$ represent the sample standard deviation

$n=9$ sample size

$\mu_o =2.3$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true mean is higher than 2,3 ppm, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.3$

Alternative hypothesis: $\mu > 2.3$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5$

The degrees of freedom are given by:

$df=n-1=9-1=8$

And the p value would be:

$p_v =P(t_{(8)}>7.5)=0.0000346$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

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3 years ago

How many 1/3 pieces can he make

Orlov [11]

He can have 2 1/3 peoces

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3 years ago