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2 years ago

What is ladder diagram of 32

1 answer:

6 0

This is what a ladder diagram looks like. 
 
2 | 32 
 2 | 16 
 2 | 8 
 2 | 4 
 2 | 2 
1
32 = 2⁵

32 / 2 = 16
16 / 2 = 8
8 / 2 = 4
4 / 2 = 2
2 / 2 = 1

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Question 2 of 10 Which describes the graph of y = (x-4)2 - 1?

WITCHER [35]

$y=(x-4)^2-1:$ $y=-\frac{4}{5}$

Sure hope this helps you

y = -4/5

4 0

1 year ago

The vertices of a quadrilateral are M(-1,1), N(1,-2),O(5,0), and P(3,3).

Komok [63]

Answer:

Quadrilateral MNOP is a rectangle.

Step-by-step explanation:

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2 years ago

X=4 in the equation e^lnx = 4. Why is this so? Can you go through the steps/rules

olga nikolaevna [1]

Answer:

e^(ln x) is just plain x

Step-by-step explanation:

The functions f(x) = e^x and g(x) = ln x are inverses of one another. In other words, one "undoes" the other.

Thus, as the rule goes, e^(ln x) is just plain x.

Here, e^(ln x) = 4 simplifies to x = 4.

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2 years ago

Help me with this please

qwelly [4]

The fraction is written as 3/4.
If you divide 3 by 4 you will get the decimal.

8 0

2 years ago

A recent study found that the average length of caterpillars was 2.8 centimeters with a

pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 2.8, \sigma = 0.7$ .

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is one subtracted by the p-value of Z when X = 4, hence:

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

$Z = \frac{4 - 2.8}{0.7}$

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0

1 year ago