All categories

3 years ago

Which of the following numbers has a value between 1.4 and 1 4/5? a. 1.34 b. 170%

1 answer:

6 0

Here, numbers are: 1.4 and 1 4/5 = 9//5 = 1.8
Now, you need to know, which number is between 1.4 and 1.8

1.34 is lower than 1.4 So, it can't
170% = 170/100 = 1.7

So, 170% is that number between 1.4 and 1 4/5

Hope this helps!

You might be interested in

If biology is 40% and chemistry is 24% and earth science is blank and so is physics. There are 4% more students signed up for an

san4es73 [151]

Given data:

The given biology percentage is B=40%.

The given chemistry percentage is C=24%.

The percenage of earth science as compare to physics is E=P+4%.

The expresion for the sum of all subjects is,

B+C+E+P=100%

40%+24%+E+P=100%

E+P=36%.............................(I)

Substitute given value in the above expression.

P+4%+P=36%

2P=32%

P=16%

The nummber of students assigned earth science is,

E=16%+4%

=20%

Thus, the percentage of the students assigned to the physics subject is 16%.

6 0

2 years ago

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

NARA [144]

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

Learn more about probability here

brainly.com/question/24756209

#SPJ4

7 0

2 years ago

an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

--------------------------------

We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem: