What is 783,264 rounded to the nearest ten thousand

Answers

Answer 1

Answer:

Answer:

780,000 if it was at least 785k then you round it to 790k

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Simplify the quotient show the work 5 divided by nine

Answers

1.8 that is the best answer

Answer:

0.56

Step-by-step explanation:

A quotient is a quantity produced by the division of two numbers.

5 / 9 = 0.555555556

0.555555556 = 0.56

Solve the equation for
X-7=13

Answers

Answer:

x = 20

Step-by-step explanation:

Isolate x by adding 7 to both sides:

x = 20

Writing Describe another methodfor solving the expression in
Exercise 1. Does this method always
work? Explain.

Answers

Answer: pontos livres

Step-by-step explanation:

o cachorrinho wwss veyry rápido

The table represents a linear function. What is the slope of the function?

Answers

Slope is the change in y over the change in x.

Using the first and last value, the change in x would be the change from -4 to 4 which is 8

The change y y would be the change from -16 to 24 which is 40

The slope would be 40/8 which reduces to 5

The answer is 5

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

Answers

Answer:

The lifetime value needed is 11.8225 hours.

Step-by-step explanation:

Problems of normally distributed samples can be 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

$Z = (X - \mu)/(\sigma)$

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 This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that $\mu = 11, \sigma = 1$ .

What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

This is the value of THE MEAN SAMPLE X when Z has a pvalue of 0.95. That is between Z = 1.64 and Z = 1.65. So we use $Z = 1.645$

Since we need the mean sample, we need to find the standard deviation of the sample, that is:

$s = (\sigma)/(√(4)) = 0.5$

So:

$Z = (X - \mu)/(s)$

$1.645 = (X - 11)/(0.5)$

$X - 11 = 0.5*1.645$

$X = 11.8225$

The lifetime value needed is 11.8225 hours.

2. Customers who purchase a certain model of car can order an engine in any of three sizes. Of all cars of this type sold, 45% have a small engine, 35% have a medium-sized engine, and 20% have a large engine. Of cars with the small engine, 10% fail an emission test within two years of purchase, while 12% of those with the medium size and 15% of those with the large engine fail. If a randomly chosen car of this model fails an emission test within two years, what is the probability it is a car with a small engine