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2 years ago

If $16 is 80% of the money I have in my wallet, how much do I have in total?

Mathematics

2 answers:

Bond [772]2 years ago

4 0

Answer:

it would be 20 dollars or 2 dollars

Step-by-step explanation:

pychu [463]2 years ago

3 0

Answer:

$20

Step-by-step explanation:

to check

16/20

divide

= .80

times 100

80%

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Evaluate. −8/125−−−−√3 A. -4/25 B. −2/5 C. 4/25 D. 2/5

larisa [96]

The correct question is: evaluate

$\sqrt[3]{- \frac{8}{125} }$

The answer is: option B. -2 / 5

Explanation:

$\sqrt[3]{- \frac{8}{125} }=- \frac{ \sqrt[3]{8} }{ \sqrt[3]{125} } =- \frac{ \sqrt[3]{2^3} }{ \sqrt[3]{5^3} } =- \frac{2}{5}$

4 0

3 years ago

For all values of x,

Keith_Richards [23]

Look at the attached picture

Hope it helps...

Good luck on your assignment..

5 0

3 years ago

I need to know this help

antoniya [11.8K]

Answer:

The slope is 2/3 and the y intercept is 5/9

Step-by-step explanation:

This is written in the form

y= mx+b where m is the slope and b is the y intercept

y = 2/3x +5/9

m = 2/3 and b=5/9

The slope is 2/3 and the y intercept is 5/9

4 0

3 years ago

For a quality control test, a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from it

amm1812

Answer:

95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

Step-by-step explanation:

We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.

The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average mpg = 15.6 mpg

s = sample standard deviation = 1.9 mpg

n = sample of minivans = 100

$\mu$ = population average mpg

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.987 < $t_9_9$ < 1.987) = 0.95 {As the critical value of t at 99 degree

of freedom are -1.987 & 1.987 with P = 2.5%}

P(-1.987 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.987) = 0.95

P( $-1.987 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.6-1.987 \times {\frac{1.9}{\sqrt{100} } }$ , $15.6+1.987 \times {\frac{1.9}{\sqrt{100} } }$ ]

= [15.22 mpg , 15.98 mpg]

Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.

7 0

3 years ago

What is the length of a side a? Round to the nearest tenth of an inch.

Galina-37 [17]

Answer:

Step-by-step explanation:

a = 16 inches

NEAREST TENTH: 20

5 0

2 years ago