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3 years ago

"what is the binary equivalent of the decimal value 97?"

1 answer:

5 0

01100001 = 97 64+32+1=97 (if the leading bit is 0 it's usually not shown)
very easy
8 bits
Bit 1 Value 128
Bit 2 value 64
Bit 3 value 32
Bit 4 value 16
Bit 5 value 8
Bit 6 value 4
Bit 7 value 2
Bit 8 value 1

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A recent survey by the cancer society has shown that the probability that someone is a smoker is P(S)=0.29. They have also deter

GuDViN [60]

<u>Answer:</u>

The correct answer option is P (S∩LC) = 0.16.

<u>Step-by-step explanation:</u>

It is known that the probability if someone is a smoker is P(S)=0.29 and the probability that someone has lung cancer, given that they are also smoker is P(LC|S)=0.552.

So using the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).

P (LC|S) = P (S∩LC) / P (S)

Substituting the given values to get:

0.552 = P(S∩LC) / 0.29

P (S∩LC) = 0.552 × 0.29 = 0.16

5 0

3 years ago

A piece of string 25 1/2 inch long will be cut into 3/4 inches pieces how many pieces will there be

weqwewe [10]

25.5/0.75 = 34 Pieces.

Hope this has helped.

6 0

3 years ago

Read 2 more answers

ava picks a number greater than 3 tens 6 ones. the number is less than 52. it has more than 6 ones. what could the. number be?

Svetlanka [38]

Any where between 37 and 51

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3 years ago

Find the area of the shape below. All measurements are in miles.

IrinaK [193]

731.04!! You multiply all together!! Mark me brainliest ?

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2 years ago

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

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

$Z = \frac{5.8-6}{0.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

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

$Z = \frac{5.8-6}{0.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

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

$Z = \frac{5-6}{0.2}$

$Z = -5$

$Z = -5$ has a pvalue of 0.

1 - 0 = 1

5 0

3 years ago