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3 years ago

Please help me with those questions

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

4 0

Answer:

5a) 2 sides are equal to each other

5b) $10\sqrt{2}$

6) 16 $\sqrt{2}$

Step-by-step explanation:

5b) I'm assuming that in this question you are trying to find the hypotenuse? So you use the pythagorean theorum, which is a^2+b^2=c^2. c represents the hypotenuse, so it would be 10^2+10^2=c^2 because the two sides are equal. Then 100+100=200 and the square root of that is 10\sqrt{2}.

6)a^2+b^2=32^2

since a=b you can write the equation as, 2a^2=1024

a^2=1024/2 , which equals a=\sqrt{512}

and simplify it to 16 $\sqrt{2}$

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A number has the same digit in its hundreds place and its hundredths place.

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Answer:

One is 2 0's before the decimal point (100) and the other is 2 decimal places AFTER the decimal point (0.01)

Essentially, the question is asking us how many times do we multiply 0.01 to get 100? This will be how many times greater 100 is compared to 0.01

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Required information Skip to question NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to re

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The result for the given 14 length bit string is-

(a) The bit strings of length 14 which have exactly three 0s is 2184.

(b) The bit strings of length 14 which have same number of 0s as 1s is 17297280.

<h3>What is a bit strings?</h3>

A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.

A null string is a bit-string with no length.

The concept used here is permutation;

ⁿPₓ = n!/(n-x)!

Where, n is the total samples.

x is the selected samples.

(a) Because there are exactly three 0s, its other digits have always been one, hence the total of permutations is equal to;

n = 14 and x = 3 digits.

¹⁴P₃ = 14!/(14-3)!

¹⁴P₃ = 2184.

Thus, the bit strings of length 14 which have exactly three 0s is 2184.

(b) Because it is a bit string with 14 digits and it can only have digits 0 or 1, we must choose 7 0s and the remaining 7 1s, hence the number possible permutations is equal to;

n = 14 and x = 7 digits.

¹⁴P₇ = 14!/(14-7)!

¹⁴P₇ = 17297280.

Thus, the bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The answer is identical to problem a, but the rest of a digits could be either 0 or 1, hence it must be doubled by 2¹¹ because there are 11 digits, each of which can be one of two options;

= 2184×2¹¹

= 4472832

Thus, he bit strings of length 14 which have at least three 1s is 4472832.

To know more about permutation, here

brainly.com/question/12468032

#SPJ4

The correct question is-

How many bit strings of length 14 have:

(a) Exactly three 0s?

(b) The same number of 0s as 1s?

(d) At least three 1s?

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4 years ago

For the 405 highway that car pass through a checkpoint, assume the speeds are normally distributed such that μ= 61 miles per hou

Strike441 [17]

Answer:

$Z = 1$

Step-by-step explanation:

Z - score

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 61, \sigma = 4$

Calculate the Z value for the next car that passes through the checkpoint will be traveling slower than 65 miles per hour.

This is Z when X = 65. So

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

$Z = \frac{65 - 61}{4}$

$Z = 1$

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