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bogdanovich [222]
3 years ago
8

If the statement "If I am hungry, then I am not happy" is assumed to be true, is its inverse, "If I am not hungry, then I must b

e happy," also always true?
Mathematics
2 answers:
elena55 [62]3 years ago
8 0
No -- other factors will influence whether your happy or not. Saying that is true is inductive reasoning, which is prone to error
andre [41]3 years ago
8 0

Answer:

Not always true

Step-by-step explanation:

The statement given is

"If I am hungry, then I am not happy"

Hungry is the cause and not happy is the result.

But hungry need not be only cause for not happy. There may be other causes such as not having money, or time, or death of some relative, or any other ailment, etc.

Hence the converse need not be true always.

So "If I am not hungry, then I must be happy,"

can be true provided all other needs are satisfied

Hence need not always be true

You might be interested in
The radius of a circle is 20 cm. Find its area in terms of π π.
natulia [17]

Answer:

A = 400\pi cm^2

Step-by-step explanation:

Area of a circle is:

A = πr²

We are given a radius of 20 cm.

A = πr²

A = π(20)²

A = π400

<em>A = 400πcm²</em>

Hope this helps.

4 0
2 years ago
Which are correct representation of the inequality minus 32X -5 less than 52 minus X select two options
Murljashka [212]

Answer:

The value of X is more than equal to -1.83.

Step-by-step explanation:

We need to find the correct representation of the inequality 'minus 32X -5 less than 52 minus X'

LHS of the inequality will be : -32x-5

RHS of the inequality will be : (52-X)

So,

-32X-5\leq 52-X

Adding X both sides

-32X-5+X\leq 52-X+X\\\\-31X-5\leq 52

Adding 5 to both sides,

-31X-5+5\leq 52+5\\\\-31X\leq 57

Dividing both sides by -31.

X\geq \dfrac{-57}{31}\\\\X\geq  -1.83

So, the value of X is more than equal to -1.83.

5 0
2 years ago
A $525 speaker now sells for 393.75 find the percent of change
asambeis [7]

Answer: The percent of change is 25%

Step-by-step explanation:

$525 - $393.75 = $131.25

131.25 (amount of change)/525 ( original amount) x 100

= 25%

5 0
3 years ago
Find the product. <br> n(n+3)
bekas [8.4K]
Distribute.
n(n+3)
Answer : n^2+3n
7 0
3 years ago
The speeds of vehicles traveling on a highway are normally distributed with an unkown population mean and standard deviation. A
Maksim231197 [3]

Answer:

The 90% confidence interval would be given by (60.09;69.91)    

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X =65 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=9 represent the sample standard deviation

n=11 represent the sample size  

2) Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=11-1=10

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,10)".And we see that t_{\alpha/2}=1.81

Now we have everything in order to replace into formula (1):

65-1.81\frac{9}{\sqrt{11}}=60.09    

65+1.81\frac{9}{\sqrt{11}}=69.91

So on this case the 90% confidence interval would be given by (60.09;69.91)    

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

3 0
3 years ago
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