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Alekssandra [29.7K]

3 years ago

13

What is the value of 5 in 43.245

1 answer:

Reptile [31]3 years ago

6 0

4➡ tens
3 ➡ ones
----------------
2➡tenths
4➡hundredths
5➡ thousandths

Anything behind the decimal point ends with a "ths" ("th" )

So, the value of "5" might be 'thousandths'

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The population of a small town can be described by the equation PN equals 4000+70n. Where n is the number of years after 2005. E

ruslelena [56]

Answer:

The current population (in 2005) is 4000 because it is the constant. The slope tells us how many people are added to the population, so 70 per year. For ex. 2008 (2008- 2005 = 3=> 3 x 70 = 210 => 210 + 4000 = 4210).

8 0

3 years ago

an employee has an annual salary of $48,700. they receive $1,530 in health insurance and $2,810 in paid time off per year. they

Anna [14]

Compensation for the year:$51,810(deducting vehicle expense)

An employee makes 48,700+1,530(health insurance)+2,810(paid time)=

53,040. Also they receive 0.53 per mile x 9,000 miles=4,770

53,040 +4,770=57,810(employment compensation)

Personal work vehicle expense: 500 x 12 months=6,000

57,810 - 6,000= 51,810

6 0

3 years ago

Read 2 more answers

X+2y=-9<br><br> Solve for Y km

kow [346]

Answer:

y = - 1/2x - 4.5

Step-by-step explanation:

Step 1:

x + 2y = - 9 Equation

Step 2:

2y = - x - 9 Subtract x on both sides

Answer:

y = - 1/2x - 4.5 Divide 2 on both sides

Hope This Helps :)

4 0

2 years ago

What is the length of AC

Alex777 [14]

Answer:

13

Step-by-step explanation:

count A up to C

6 0

2 years ago

Using traditional methods it takes 92 hours to receive an advanced flying license. A new training technique using Computer Aided

DanielleElmas [232]

Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

$H_0: \mu = 92$

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

$H_1: \mu > 92$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

92 is tested at the null hypothesis:

This means that $\mu = 92$

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that $n = 70, X = 93, \sigma = \sqrt{36} = 6$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}$

$z = 1.39$

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

8 0

2 years ago