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3 years ago

What is -32 divided by 9

Mathematics

2 answers:

soldier1979 [14.2K]3 years ago

8 0

The answer is −3.555555555555556 but you can just put -3.55

kolezko [41]3 years ago

8 0

Answer:

Reduce the expression, if possible, by cancelling the common factors.

Exact Form: -32/9

Decimal Form: -3.5

Mixed number form: -3 5/9

Step-by-step explanation:

1. Because you are dividing, are the signs are different, we know that the signs will be different. Begin with just writing it will be negative. Get it out of the way.

2. Then proceed to divide like normal. 9 goes into 32 3 times unevenly, so we that will be the whole number.

3. Now, find the difference between the number 27 (the last number we were able to multiply 9 by) and 32. That will be the numerator.

4. Use the original denominator, in this case, the dividend, 9.

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In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316$

$0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524$

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32$

And rounded up we have that n=335

4) Part d

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599$

And rounded up we have that n=649

5 0

3 years ago

Me not being a complete id.iot-

Aleks04 [339]

Answer:

im pretty sure the answer is d

Step-by-step explanation:

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2 years ago

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The shipping would be 15.75 because 5% of 15 is 0.75, so add 15 to 0.75 and the answer is 15.75 :)

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3 years ago

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madam [21]

Answer:

$501,049.37

Step-by-step explanation:

For computing the amount after 22 years we need to applied the future value which is shown in the attachment below:

Given that

PMT = $9,000

NPER = 22 years

Annual rate = 0.078

Quarterly= 0.078 ÷ 4 = 0.0195

Effective annual rate = (1.0195^4) - 1 = 0.0803113041

Now applied the formula which is given below

= -FV(RATE;NPER;PMT;PV)

After applying the above formula, the future value is $501,049.37

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3 years ago

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allsm [11]

Answer:

Step-by-step explanation:

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3 years ago

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