What is 173% of 110? Round to the nearest hundredth

Please help i really need it, i will give brainliest to the right answers

Dmitry_Shevchenko [17]

Answer:

-292 1/2

Step-by-step explanation:

PEMDAS

Perenthesis

2 To the power of 4 = 16 + 1/4 = 16/1 + 1/4 X -18

4 0

2 years ago

Tramserran? Do you think you could lend a hand for these questions? You have saved my life on more than one occasion. It's Algeb

Makovka662 [10]

Answer:

#1). 6 , 18 , 54

#2). 5/3 , 14 / 9 , 41/27

#3). 1.5 , 2.5 , 2.5

Step-by-step explanation:

#1).

g(x) = 3x

g(2) = 3 . 2 = 6

g²(2) = 3 . 3 . 2 = 18

g³(2) = 3 . 3 . 3 . 2 = 54

#2).

g(x) = 1/3 x + 1

g(2) = 2/3 + 1 = 5/3

g²(2) = g(5/3) = 5/9 + 1 = 14/9

g³(2) = g(14/9) = 14/27 + 1 = 41/27

#3).

g(x) = -1 ║x - 2 ║ + 3

g(0.5) = -1 (1.5) + 3 = 1.5

g²(0.5) = g(1.5) = -1(0.5) + 3 = 2.5

g³(0.5) = g(2.5) = -1(0.5) + 3 = 2.5

8 0

3 years ago

Read 2 more answers

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard

avanturin [10]

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

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:

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

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that $\mu = 33208, \sigma = 2503$ .

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when $X = 33208+633 = 33841$ subtracted by the pvalue of Z when $X = 33208 - 633 = 32575$

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57$

So

X = 33841

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

$Z = \frac{33841 - 33208}{357.57}$

$Z = 1.77$

$Z = 1.77$ has a pvalue of 0.9616

X = 32575

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

$Z = \frac{32575- 33208}{357.57}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

4 0

3 years ago

A long distance runner starts at the beginning of a trail and runs at a rate of 4 miles per hour. One hour later, a cyclist star

Scrat [10]

T = 1/4 hour will be the answer

7 0

3 years ago

........................................

Sergio [31]

What is your Question...

3 0

3 years ago