What is the base 10 representation of 11102?

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

The base two number one one one zero is equal to one times eight, plus one times four, plus one times two, plus zero times one, which simplifies to fourteen.

Answer 2

Answer:

Answer:

1.1102 * 10^4

Step-by-step explanation:

11102

= 1.1102 * 10^4

Related Questions

From 1985 to 2007, the number B B of federally insured banks could be approximated by B ( t ) = − 329.4 t + 13747 B(t)=-329.4t+13747 where t is the year and t=0 corresponds to 1985. How many federally insured banks were there in 1990?
Question is in Screenshot/Attachment! Hurry pls.
What is nearest tenth of 0.273
A quadrilateral has 4right angles and 2 pair of opposite sizes that are parallel. What quadrilateral could be
9/8+7/40= and does the answer simplify

you drink a beverage with 120 mg of caffeine. Each hour, the amount m of caffine in a persons system decreases by 12%. About how much caffeine will be in your system after 3 hours? Round your answer to the nearest milligram

Answers

Answer:43.2

Step-by-step explanation: multiply 120 x 0.12 and you get 14.4, since it’s 3 hours multiply 14.4 3 times and u get 43.2

5. What is the value of x if the quadrilateral is a kite? B X+2 C С 13 A Xth12 D

Answers

Answer: just had this problem! X = 11

According to a random sample taken at 12 A.M., body temperatures of healthy adults have a bell-shaped distribution with a mean of 98.28degreesF and a standard deviation of 0.63degreesF. Using Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2 standard deviations of the mean? What are the minimum and maximum possible body temperatures that are within 2 standard deviations of the mean? At least nothing% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.

Answers

Answer:

At least 75% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.

The minimum possible body temperature that is within 2 standard deviation of the mean is 97.02F and the maximum possible body temperature that is within 2 standard deviations of the mean is 99.54F.

Step-by-step explanation:

Chebyshev's theorem states that, for a normally distributed(bell-shaped )variable:

75% of the measures are within 2 standard deviations of the mean

89% of the measures are within 3 standard deviations of the mean.

Using Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2 standard deviations of the mean?

At least 75% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.

Range:

Mean: 98.28

Standard deviation: 0.63

Minimum = 98.28 - 2*0.63 = 97.02F

Maximum = 98.28 + 2*0.63 = 99.54F

The minimum possible body temperature that is within 2 standard deviation of the mean is 97.02F and the maximum possible body temperature that is within 2 standard deviations of the mean is 99.54F.

Check Your Understanding!1. A Ford Escape has a 14-gallon fuel tank and can travel approximately 26 miles with one gallon of
gas, Write a function d(x), that gives the distance the Escape can travel with x gallons of gas in
the tank.

Answers

Given :

Maximum capacity of fuel tank , M = 14 gallon .

It can travel approximately 26 miles with one gallon of gas.

To Find :

A function d(x), that gives the distance the Escape can travel with x gallons of gas in the tank.

Solution :

Car can travel 26 miles in one gallon of gas.

So , distance covered in x gallon of gas :

$D(x)=26x$

( Here, 0 ≤ x ≤ 14 ) or x ∈ [ 0 , 14 ] .

Hence , this is the required solution.

A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean

Answers

Answer:

The minimum sample size needed is $n = ((1.96√(\sigma))/(4))^2$ . If n is a decimal number, it is rounded up to the next integer. $\sigma$ is the standard deviation of the population.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1 - 0.9)/(2) = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

$M = z(\sigma)/(√(n))$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean?

A sample of n is needed, and n is found when M = 4. So

$M = z(\sigma)/(√(n))$

$4 = 1.96(\sigma)/(√(n))$

$4√(n) = 1.96√(\sigma)$

$√(n) = (1.96√(\sigma))/(4)$

$(√(n))^2 = ((1.96√(\sigma))/(4))^2$

$n = ((1.96√(\sigma))/(4))^2$

The minimum sample size needed is $n = ((1.96√(\sigma))/(4))^2$ . If n is a decimal number, it is rounded up to the next integer. $\sigma$ is the standard deviation of the population.

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 139 millimeters, and a variance of 49. If a random sample of 34 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 1.8 millimeters

Answers

The probability that the sample mean would differ from the population mean by more than 1.8 millimeters is approximately 0.0668.

What is the standard deviation?

A standard deviation (σ) is a measure of the distribution of the data in reference to the mean.

The standard deviation of the population is $√(49) = 7$ millimeters. The standard error of the sample mean is then

$(7)/(√(34)) = (7)/(5.874) \approx 1.2$ millimeters.

The probability that the sample mean would differ from the population mean by more than 1.8 millimeters is the probability that it falls outside of the interval $(139 - 1.8, 139 + 1.8)$ . We can use the standard normal distribution to approximate this probability.