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3 years ago

Rationalize the denominator of the fraction and enter the new denominator below.8/2-√12

2 answers:

Studentka2010 [4]3 years ago

8 0

$\frac{8}{2-\sqrt{12}}=\frac{8}{2-\sqrt{4\cdot3}}=\frac{8}{2-2\sqrt3}=\frac{8}{2(1-\sqrt3)}=\frac{4}{1-\sqrt3}\\\\\frac{4}{1-\sqrt3}\cdot\frac{1+\sqrt3}{1+\sqrt3}=\frac{4(1+\sqrt3)}{1^2-(\sqrt3)^2}=\frac{4(1+\sqrt3)}{1-3}=\frac{4(1+\sqrt3)}{-2}=-2(1+\sqrt3)}\\\\=-2-2\sqrt3$

$if\ \frac{8}{2}-\sqrt{12}=4-\sqrt{4\cdot3}=4-2\sqrt3$

MakcuM [25]3 years ago

7 0

$\frac{8}{2-\sqrt{12}}\\
\frac{8(2+\sqrt{12})}{4-12}=\\
\frac{8(2+\sqrt{12})}{-8}=\\
-(2+\sqrt{12}})=\\
-2-\sqrt{12}=\\-2-2\sqrt3
$

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withdraws $ from bank account once each day for days. What integer represents the change in the amount in the account? Use pen

REY [17]

Answer:

a. Sam withdraws $11 from his account. That means his account balance reduces by $11 so the integer is -$11.

He does this 4 times so;

= 4 * -11

= -$44

b. He then deposits $11 once every day for 4 days.

= 4 * 11

= $44

c. The integer for withdrawals is a negative figure to show that the balance was decreasing. The Integer for deposits is positive to show that the balance was increasing.

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3 years ago

What is 8 thousandths less than 4.989?

emmainna [20.7K]

8000-4.989 so when you see the word less than it means that you have to subtract and more than you have to add so the answer for this question will be 7,995.011

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2 years ago

A population of 1,000 squirrels lives on a rectangle-shaped farm plot that is 25 feet long by 10 feet wide what is the populatio

Lemur [1.5K]

First, find the area of the plot:
25*10=250

1000/250 is 4

the population density is 4 squirrels per square foot.

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3 years ago

Which of the following is fartherest to the left on a number line 1/5, (0.1)to second power, 0.02

Alexxx [7]

1/5 is .2
.1^2 is .01
.02 is .02

So the furthest left would be the smallest number, which in this case is 0.1^2

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4 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago